Practice: A sample of copper absorbs 35.3 kJ of heat, which increases the temperature by 25 degrees Celsius, determine the mass (in kg) of the copper sample if the specific heat capacity of copper is 0.385 J/g *C.

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Both **heat capacity (C)** and **specific heat capacity (c) **deal with the amount of heat required to change a compound’s temperature by 1 Kelvin. Specific heat capacity (c) deals with changing 1 gram of a compound.

By using **calorimetry** we can measure the thermal energy of a reaction (system) by measuring the change in heat of the surroundings.

Concept #1: Heat capacity vs. Specific Heat Capacity

**Transcript**

Welcome back guys! In this new video, we're going to take a look at calorimetry under a constant volume.

Now first we're going to say that every object has its own heat capacity. And just remember heat capacity uses a capital C, and this is just the amount of heat that's required to change an object's temperature by 1 Kelvin (K). Here we're just going to say heat capacity is simply heat/the change in temperature, so Final - Initial. And we're going to say the units are usually in Joules (J)/Kelvin or kiloJoules/Kelvin.

q J

C = ∆T [ in units of K ]

Sometimes your professor may even do it over Degrees Celsius, so just look to see what units were they want the answer to be in. That's what you always have to make sure so that you get the correct answer.

Now similar to heat capacity, there's another property. This property is known as Specific Heat Capacity, and notice that this one is lower case c, and this is the amount of heat that's required to change 1g of substance by 1K.

So here we're just going to introduce the concept of mass, and mass here would be in grams. So it's similar to heat capacity but more in depth, where talks about mass. Now what we're going to say here is if we know the specific heat capacity of a substance, we can rearrange this formula here so that we can solve for the amount of heat absorbed or released. So if we want to rearrange this equation up here, so C = q/m * change in temperature.

q

C = m * ∆T

Just multiply both sides by mass * change in temperature.

q

m * ∆T (C) = m * ∆T m * ∆T

And then you say that q = mcAT.

q = mc∆T

So just remember q = mcAT, just read the delta sign, the triangle as an A. So that's the easy way for you to remember the equation, q = mcAT. This is the equation that we use anytime they give us specific heat capacity, and we know that change in temperature, and they give us the mass of an object.

Now based on the concept we've just looked at let's take a look at this Question here. It says:

At constant volume, the heat of combustion of a particular compound is -4621.0 kJ/mol. We say when 2.319 grams of this compound (molar mass = 192.75 g/mol), remember it has a molar mass that I give to you, was burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 3.138 oC. What is the heat capacity of the calorimeter in J/K?

First of all, this image that we have here is called a bomb calorimeter. It's a way of us actually figuring out how much energy does an object contain in it. The way it works is, we have our sample put inside to the small little bowl here, and that is placed inside of those water. This whole unit is referred to as a bomb. And what's going to happen is, we're going to actually use these rods here, and we're going to send an electrical current, and it's actually going to cause this sample to explode inside of the bomb calorimeter. So that's why it's called the bomb calorimeter because we actually cause the contents to explode. And when they explode, they're going to give off some heat. Now there's a temperature in the water, this temperature is actually going to record how much the water temperature increases by. In that way, we can figure out how much heat that that sample release and from that we can figure out how many calories it has, how much heat it started with, we could find out a lot of information. This is the way to do certain food sciences, to figure out the calories we have in a food. They use similar technique such as this.

Now we talked about Joules as being the standard type of energy for heat, but remember you could also have instead of Joules for q you could have kiloJoules, but you could also have calories, large calories, these are the calories that you see in food, and kilowatt-hours. So these are other units, energy conversion factors that we can use in place of Joules for q. Now we're not going to work on converting from Joules to calories or kilowatt-hours, I just want to give you guys these tables so that you can see all the different types of energy conversions that you should know for lecture.

Now let's take a look at this question, we want to calculate heat capacity. Remember this is capital C, and so heat capacity = q/change in temperature. And what we need to realize here is, we need to isolate Joules, we need to find Joules for q, and we need the temperature to be in Kelvin. We already have half of it right off the bat, the easy part. The easy part here is I told you that the temperature rose by this much.

Example #1: Bomb Calorimeter

**Transcript**

That right there represents our change in temperature, but the thing is we just need to change those oC into K. And how do we do that? Remember the goal from oC to K, you just add 273.15 to it, when we do that we get 276.288 K.

3.138 oC + 273.15 = 276.288 K

Now what we have to do is we have to isolate q, we have to find Joules. Now the closest place we're going to find Joules are right here in kiloJoules. So we need to find a way of isolating those kiloJoules there. Now the only way I can isolate those kiloJoules is if I find a way to cancel out these moles that we have. And the way I'm going to cancel out those moles is if I convert these grams that I gave to you into moles and multiply times this whole thing. So what we're going to do is we're going to start out with 2.319 g, now we can change this into moles because I gave you the molecular weight, the molar mass of the object. So we're going to say for every 1 mole of this substance it's 192.75 g, grams cancel out and we've just isolated our moles.

1 mol

2.319 g x 192.75 g

I'm going to take myself out of the image so that we'll have more room to work with guys. So we have moles now, now I can just multiply times the heat of combustion, so this would be -4621.0 kJ on top, 1 mol on the bottom. Moles cancel out and now I have kiloJoules isolated.

1 mol -4621.0 kJ

2.319 g x 192.75 g x 1 mol

But remember I don't want kiloJoules, I want Joules. So one more step, kiloJoules go on the bottom, Joules on the top. Kilo is a metric prefix, so for every 1 kilo it's 103. When we work all of that out that gives me -55 600 J, so that's my q. So I'm going to plug it up here. So divide that by the temperature we just isolated and you'll get -201 J/K.

1 mol -4621.0 kJ 10J

2.319 g x 192.75 g x 1 mol x 1 kJ

= - 55 600 J

So I'm going to plug it up here. So divide that by the temperature we just isolated and you'll get -201 J/K.

q

C = ∆T

= - 55 600 J

276.288 kJ

= - 201 J/K

Now this question seems like a lot of information into this, but the thing is just focus on what they're asking us to find, they wanted us to find heat capacity. Heat capacity is just simply q/change in temperature, we knew change in temperature immediately so we just convert it that to Kelvin and then you're now to work on isolating the one variable you needed, you needed kJ, you need to isolate kJ. We isolated that by changing the grams I gave you into moles and then multiplying times the heat of combustion. Now when they say heat of combustion, remember heat means q, but more importantly here it means delta H. They could have said that enthalpy of combustion, also that be the same exact thing. Okay so just remember heat and enthalpy is the same thing.

A **bomb calorimeter** measures the amount of calories within a substance through combustion. In other words, we “blow it up” and measure the amount of heat it releases.

If you are given energy or heat, with specific heat capacity and mass then you will most likely use **q = mcΔT**.

Example #2: In an experiment a 9.87 carat (1 carat = 0.200g) diamond is heated to 72.25^{o}C and immersed in 22.08 g of water in a calorimeter. If the initial temperature of the water was 31.0^{o}C what is the final temperature of the water? (c_{diamond} = 0.519) (c_{water} = 4.184 ).

Practice: A sample of copper absorbs 35.3 kJ of heat, which increases the temperature by 25 degrees Celsius, determine the mass (in kg) of the copper sample if the specific heat capacity of copper is 0.385 J/g *C.

When dealing with heat it is important to remember that if one compound is gaining heat that means another compound is losing heat.

Practice: 50.00 g of heated metal ore is placed into an insulated beaker containing 822.5 g of water. Once the metal heats up the final temperature of the water is 32.08 degrees Celsius. If the metal gains 14.55 kJ of energy, what is the initial temperature of the water?

The enthalpy or heat of a reaction can be calculated through the use of a coffee cup calorimeter.

Concept #2: Coffee Cup Calorimeter

**Transcript**

Welcome back, guys! In this new video, we're going to take a look at chemical reactions when the calorimetry is done under constant pressure. What you’re going to see here is this image right here. We're going to say that this is called a coffee calorimeter. We're going to say it’s used basically to help us find the enthalpy of a reaction. Remember, enthalpy just means delta H. The coffee cup calorimeter has certain parts to it. Here we have the stirrer, which just helps us to mix the water that surrounds our sample. Here we’re going to have our thermometer, which measures the temperature change involved when the sample undergoes its chemical reaction. The sample was going to be releasing heat. The water is going to be absorbing that heat. The thermometer is just to calculate the change in temperature that the water undergoes. Here we’re going to have the two styrofoam cups that are placed together. They work as insulation. That way, no heat is lost to the outside environment. All the temperature or energy changes that occur are going to happen between the water and the sample. Here we're going to have our water. Our water acts as the surroundings. Here we're going to have our sample, which is going to undergo the chemical reaction. This represents our system.

Remember, under constant pressure it's possible for us to calculate the enthalpy of the reaction through the use of this coffee cup calorimeter.

Concept #3: Calculating the Heat of the Solution

**Transcript**

Now let's take a look at it this example question. Here it says you place 50 ml of 0.100 molar NaOH into a coffee cup calorimeter at 50 degrees Celsius and carefully add 75 mls of 0.100 molar hydrochloric acid also at 50 degrees Celsius. After stirring, the final temperature of the solution is 76.12 degrees Celsius. Heat capacity and the density of water are also given.

From this, we have to calculate two parts. For part A, we have to figure out what is the heat or Q of the solution in joules. Since we're going to need room to do both of these questions, I'm going to take myself out of the image guys, so we have more room to work with.

What we should realize is that we need to find the heat of the solution. The problem is it's hard to do it directly. But we can find the Q of the water. If we know Q of water, we can then say Q of water equals negative Q of the solution. Remember, in this process, the sample will undergo a chemical reaction and release heat that's why it’s negative. The water will absorb that heat that's why it's positive. We can find the Q of water because we can say Q of water equals mCΔT. We already know the specific heat water. I gave it to you. I said it’s 4.184 joules over degrees times Celsius. Our temperature change, our final temperature is 76.12 degrees Celsius. Our initial for both samples when we mix them together were 50 degrees Celsius.

The harder part is just figuring out what the mass of the water is. But we can figure this out because when we say 50 ml and 75 ml, those two things actually represent the volume of water. What's really going on here that we have that many milliliters of water and dissolved within it are those moles of NaOH and HCl. We're going to take those 50 mls and that 75 ml and add them together to give me 125 mls of water. Since I know the density of water here, I can use that to find the grams of water. Multiply this times the density. Mls cancel out and I have 125 grams of water, which I can just plug into here.

When I multiply everything out, that gives me 13,660.8 joules as the Q of water. We can just plug it into here so we’re going to have 13,660.8 joules equals negative Q of solution. But I want Q of solution, not negative Q of solution. So then all you have to do is divide both sides by -1. Now, Q of solution equals negative 13,660.8 joules as your final answer. Here we’re not concerning ourselves with number of significant figures because here we’re just trying to figure out what the answer is. If you will get this type of question on your exam, your professor would specify do they want sig figs or not.

The **heat of the solution** can be determined by first determining the **heat of water**.

Concept #4: Calculating the enthalpy of the reaction

**Transcript**

For B, we have to calculate the enthalpy, delta H, in joules per mole for the formation of water. Here we have to figure out the enthalpy of water. We're going to say delta H of reaction equals Q of solution divided by the moles of whatever they're asking us to find. Here, they’re asking for the formation water so it’s moles of water. We already have half of this because we just figured out what the Q of solution was, so we’ll just plug it in. Now what we have to do is we have to figure out how many moles of water we have.

What we should realize here is I give you information on NaOH and HCl. Remember, the word of when it’s in between two numbers means multiply. Remember, when I say big M that means molarity. Molarity equals moles over liters. When I say 0.100 molar NaOH, that really means 0.100 moles of NaOH over 1 liter. We always assume it's always moles over liters.

If you guys don’t remember this too well, go back to a few videos past where we talked about molarity and molarity in stoichiometry because to do this question, it's essential that you remember those steps from earlier. We’re going to work it out down here. Since I gave you information on two reactants and I'm asking you how much product do you have in moles, this is really a limiting reactant type of question. We're going to have to first do it with the molarity of NaOH, see how many moles of water that gives us then do the same thing with HCl. Since this is a limiting reactant type of question, the one that gives us a smaller amount of moles of water, those are those moles we’re going to use. We have 50 ml of 0.100 molar NaOH. I need to change my ml into liters. 1 ml on the bottom, 10 to the negative 3 liters on the top. Mls cancel out and I finally isolate liters. Now that I have liters, I can multiply it times the molarity to find the moles of NaOH. We need the moles of water though. We need to go from these moles I've given that we just found to our moles of unknown. Remember when we do that jump, we do a mole to mole comparison. We look at the coefficients in the balanced equation. According to our balanced equation, for every one mole of NaOH, we have one mole of water. NaOH on the bottom and mole of H2O what we’re looking for in the top. Since this is in 10 to the negative 3, put it in brackets. When we work it out, we get 0.005 moles of water.

Now we have to do the same thing with HCl. We have 75 ml of 0.100 molar HCl. Same thing, we change the mls into liters first then we multiply it times the molarity. For every one mole of HCl, we have one mole of H2O. Liters cancel out, moles cancel out. We have 0.0075 moles of water. Remember, since this is the limiting reactant type of question, we're going to go with the smaller moles of water. Those are the moles we're going to plug down here to find our enthalpy. When we plug that in, we get back negative 2.73 times 10 to the 6 joules per mole.

This question had a lot of parts to it. It became even harder because we had to incorporate concepts that we learned a few videos back. Just because we passed molarity and stoichiometry involved with them doesn't mean we don't have to see it again this semester. Those types of ideas always resurface at some point. We had to use them in order to solve this question. Remember, if we’re at constant pressure, that's when we use a coffee cup calorimeter. Its whole purpose is to help us find the enthalpy of a reaction. Remember, if your enthalpy, I may not have said that but if your enthalpy is positive, that means you’re absorbing energy or absorbing heat so you're endothermic. If you're releasing heat or releasing energy by the system, then it's exo. It’s negative delta H. Just remember the concepts that we covered here and the approaches we take. When you see a question like this, you'd be better prepared to do it.

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Concept #1: Heat capacity vs. Specific Heat Capacity

Example #1: Bomb Calorimeter

Example #2: In an experiment a 9.87 carat (1 carat = 0.200g)...

Practice #1: A sample of copper absorbs 35.3 kJ of heat, whi...

Practice #2: 50.00 g of heated metal ore is placed into an i...

Concept #2: Coffee Cup Calorimeter

Concept #3: Calculating the Heat of the Solution

Concept #4: Calculating the enthalpy of the reaction

A piece of iron (mass = 100.0 g) at 398 K is placed in a Styrofoam coffee cup calorimeter containing 25.0 mL of water 298 K. Assuming that no heat is lost to the cup, what will be the final temperature of the water? The specific heat capacity of iron = 0.449 J/gºC.
A. 308 K
B. 328 K
C. 338 K
D. 368 K
E. 388 K

If three samples of silver, one with a mass of 10.0 grams, another with a mass of 50.0 grams and a third with a mass of 100.0 grams each absorb 36.3 kJ of heat which sample will experience the greatest increase in temperature?
The 10.0 gram sample
The 50.0 gram sample
The 100.0 gram sample
Each will have the same increase in temperature
Depends on the starting temperature of each

The specific heat of copper metal is 0.385 J/g K. How many joules of heat are necessary to raise the temperature of a 1.42-kg block of copper from 25.0 °C to 88.5 °C?
A) 3.47 x 104 J
B) 34.7 J
C) 2.34 x 105 J
D) 8.46 J

Which substance (with specific heat capacity provided) would show the smallest temperature change upon gaining 200.0 J of heat?
a. 50.0 g Al, CAl = 0.903 J/g°C
b. 50.0 g Cu, CCu = 0.385 J/g°C
c. 25.0 g granite, Cgranite = 0.79 J/g°C
d. 25.0 g Au, CAu = 0.128 J/g°C
e. 25.0 g Ag, CAg – 0.235 J/g°C

A sample of 1.67 grams of compound Y is burned completely in a bomb calorimeter which contains 2500 g of water. The temperature rises from 24.273ºC to 24.587ºC. What is ∆Urxn for the combustion of compound Y? The hardware component of the calorimeter has a heat capacity of 3.29 kJ/ºC. The specific heat of water is 4.184 J/g·ºC, and the MW of Y is 117 g/mol.
1. -344.7
2. -615.4
3. -392.1
4. -322.0
5. -302.5
6. -185.4
7. -482.0
8. -652.5
9. -278.2
10. -424.3

The same reaction in a bomb and coffee-cup calorimeter:
a. will give the same value of Hrxn because it is the same reaction.
b. will give the same value for Hrxn because both systems are identical.
c. will give the same values because both systems are at constant temperature.
d. will give slightly different values because the coffee-cup calorimeter will do some PV work.
e. will give slightly different values because the bomb calorimeter will do some PV work

A total of 2.25 moles of a compound are allowed to react with water in a foam coffee cup and the reaction produces 83.1 g of solution. The addition of the compound caused the temperature of the solution to increase from 20.5 oC to 32.1 oC. What is the enthalpy of the reaction? Assume no heat is transferred or lost to the surroundings or to the foam coffee cup. The specific heat of the solution is 4.184 J/(g×oC) .

The molar heat capacity of silver is 25.35 J/mol × oC . Calculate how much energy (in kJ) it would take to raise the temperature of 15.7 g of the silver metal by 17.2 oC.

Which of the following substances (with specific heat capacity provided) would show the greatest temperature change upon absorbing 100.0 J of heat?
a) 10.0 g Ag, CAg = 0.235 J/g°C
b) 10.0 g H2O, CH2O = 4.18 J/g°C
c) 10.0 g ethanol, Cethanol = 2.42 J/g°C
d) 10.0 g Fe, CFe = 0.449 J/g°C
e) 10.0 g Au, CAu = 0.128 J/g°C

How much energy must be transferred to raise the temperature of a cup of coffee that is 250 mL from 20.5 °C to 368.8 K? Assume that water and coffee have the same density 1.00 g/mL and specific heat capacity (4.184 J/g•K).

Which substance (with specific heat capacity provided) would show the smallest temperature change upon gaining 200.0 J of heat?
a) 50.0 g Al, CAl = 0.903 J/g°C
b) 50.0 g Cu, CCu = 0.385 J/g°C
c) 25.0 g granite, Cgranite = 0.79 J/g°C
d) 25.0 g Au, CAu = 0.128 J/g°C
e) 25.0 g Ag, CAg = 0.235 J/g°C

When a fresh breath of air is drawn into the lungs, it is heated by your body. Let’s assume it reaches thermal equilibrium with your body temperature of 37°C. Given the temperature, pressure, and the average molar mass of air, you could easily calculate that your breath of 0.48 L corresponds to 0.51 g of air. The specific heat of air is 1.0 J/g•°C.
I. Given that the temperature outside is 21°C, how much heat is required to take the air in your 0.48 L breath from 21°C to 37°C?
II. The heat calculated in the previous question is lost every time you exhale. Assuming 15 breaths per minute, how much heat would be lost in one day by exhaling? Furthermore, assuming a daily energy intake of 8400 kJ (corresponding to 2000 food Calories), what percent of your daily energy intake is lost as heat due to exhaling during a 24 hour period?

Adding water to a steel pan on a flame slows the rise in temerature of the pan, which of the following concepts does this represent?
A. Heat capacity is an intensive property
B. Heat capacity is an extensive property
C. The example does not relate to heat capacity
D. Temperature is a state function

If it takes 0.216 kJ of heat to raise the temperature of a 12.0 g piece of Al from 15.5 to 35.5 oC, what is the specific heat (J/g K) of Al?

How much energy is required to change the temperature of 21.5 g Cu from 27 to 88.1 °C? The specific heat capacity of copper is 0.385 J/gK.
223 J
506 J
641 J
729 J
3.41 x 103

In a bomb calorimeter, reactions are carried out at:
(A) constant pressure
(B) constant volume
(C) 1 atm pressure and 25°C
(D) 1 atm pressure and 0°C

When 68.00 J of energy are added to a sample of Gallium that is initially at 25°C, the temperature rises to 38.0° C. What is the volume of the sample?
The specific heat of Gallium is 0.372 J•g -1•°C-1.
Density of Gallium is 5.904 g•cm–3.
a) 2.38 cm3
b) 4.28 cm3
c) 14.1 cm3
d) 31.0 cm3

A student mixes 100 mL of 0.50 M NaOH with 100 mL 0f 0.50 M HCl in a Styrofoam® cup and observes a temperature increase of ΔT1. When she repeats this experiment using 200 mL of each solution, she observes a temperature change of ΔT2. If no heat is lost to the surroundings or absorbed by the Styrofoam® cup, what is the relationship between ΔT1 and ΔT2?
a) ΔT2 = 4 ΔT1
b) ΔT2 = 2 ΔT1
c) ΔT2 = 0.5 ΔT1
d) ΔT2 = ΔT1

More heat is derived from cooling one gram of steam at 100°C to water at 50°C than from cooling one gram of liquid water at 100°C to 50°C because
a) the steam is hotter than the water.
b) the steam occupies a greater volume than the water.
c) the density of water is greater than that of steam.
d) the heat of condensation is evolved.

Three separate 3.5g blocks of Al, Cu, and Fe at 25 °C each absorb 0.505 kJ of heat. Which block reaches the highest temperature? The specific heats of Al, Cu, and Fe are 0.900 J/g•°C, 0.385J/g•°C, and 0.444 J/g•°C, respectively.
a) Fe
b) Fe and Cu
c) Al and Cu
d) Al
e) Cu

When a 45.0 g sample of an alloy at 100.0 °C is dropped into 100.0 g of water at 25.0 °C, the final temperature is 37.0 °C. What is the specific heat of the alloy? The specific heat of water is 4.184 J•g-1•°C-1
a) 0.423 J•g-1•°C-1
b) 1.77 J•g-1•°C-1
c) 9.88 J•g-1•°C-1
d) 48.8 J•g-1•°C-1

The molar heat capacity of silver is 25.35 J/mol•°C. Calculate how much energy (in kJ) it would take to raise the temperature of 15.7 g of the silver metal by 17.2 °C.

Two solids of equal mass, labeled X and Y, are placed in contact with each other. Solid X has an initial temperature of 100°C and Solid Y has an initial temperature of 25°C. After some time, both solids are at 40°C. Which statement is correct regarding the direction of heat transfer and specific heat capacity (Assume heat transfers only between X and Y):
Direction of heat transfer Specific heat capacity
A. From X to Y X has greater heat capacity
B. From X to Y Y has greater heat capacity
C. From Y to X X has greater heat capacity
D. From Y to X Y has greater heat capacity

A 19.78 g piece of nickel (specific heat of nickel = 0.444 J/g oC) was heated to 103.5 oC and then plunged into a beaker containing 87.9 grams of water at 14.7 oC. Determine the final temperature of the metal after the system attains thermal equilibrium. (the specific heat of water is 4.184 J/g oC)
A. 59.1 oC
B. 44.4 oC
C. 16.8 oC
D. 12.5 oC
E. 55.6 oC

Determine the specific heat capacity of an alloy that requires 59.3 kJ to raise the temperature of 150.0 g alloy from 298 K to 398 K.
A) 3.95 J/g°C
B) 1.87 J/g°C
C) 2.29 J/g°C
D) 2.53 J/g°C
E) 4.38 J/g°C

Suppose a 50.0 g block of silver at 100°C is placed in contact with a 50.0 g block of iron at 0°C, and the two blocks are insulated from the rest of the universe. The final temperature of the two blocks
A) will be higher than 50°C.
B) will be lower than 50°C.
C) will be exactly 50°C.
D) is unrelated to the composition of the blocks.
E) cannot be predicted.

When 75.4 J of energy is absorbed by 0.25 mol of CCl 4, what is the temperature change of CCl4? The specific heat capacity of CCl 4 is 0.861 J/g·°C. Molar mass of CCl 4 is 153.81 g/mol.
A) 17.8°C
B) 21.9°C
C) 2.3°C
D) 9.1°C
E) 44.6°C

Which of the following statements are correct?
A. specific heat capacity is defined as the amount of heat energy required to raise the temperature of an object 1K
B. under conditions of constant pressure, no work is done on or by the system
C. molar heat capacity is defined as the amount of heat energy required to raise the temperature of 1 mole of a substance 1K
D. heat is always transferred from an object with a lower temperature to an object with a higher temperature

In a calorimetry experiment a student determines the enthalpy for the reaction of magnesium with hydrochloric acid to be -430 kJ/mol. The true enthalpy for this reaction is -450 kJ/mol. Which of the following sources of error could account for this discrepancy?a. After recording the mass, some of the magnesium was dropped on the counterb. Some of the hot hydrogen gas escaped during the reaction.c. The calorimeter was not tightly shut for the experiment.d. The final temperature was recorded before the reaction was complete.e. All of these errors would cause a falsely high value for the enthalpy

A bar of hot metal is placed in water in an insulated container and the two are allowed to reach thermal equilibrium. When 1.0 kg of metal at 100°C is placed in 2.0 kg of water, the temperature water bath raises from 20°C to 25°C. What is the specific heat capacity of the metal (J/g K)?a) 0.5b) 1.5c) 0.22d) 25e) 0.025

The two aqueous solutions are not at room temperature and are then mixed in a coffee cup calorimeter. The reaction causes the temperature of the resulting solution to fall below room temperature. Which of the following statements is TRUE? a. The products have a lower potential energy than the reactants.b. This type of experiment will provide data to calculate ΔE rxnc. The reaction is exothermic.d. Energy is leaving the system during the reaction.e. None of the above statements is true.

In constant-volume calorimetry, what is FALSE?a. There is no work performed.b. The total energy is measured by the change in pressure.c. The total energy is determined only by heat.d. The heat capacity of the calorimeter needs to be known in order to determine the heat.e. All of the above are true.

It takes 75.0 J to raise the temperature of an 19.3 g piece of unknown metal from 21.0oC to 46.7oC. What is the specific heat of the metal?

A 55.0-g piece of metal is heated in boiling water to 99.8 °C and then dropped into cool water in an insulated beaker containing 225 mL of water with an initial temperature of 21.0 °C. The final temperature of the metal and water is 23.1 °C, what is the specific heat capacity of the metal?

Two aqueous solutions are both at room temperature and are then mixed in a coffee cup calorimeter. The reaction causes the teperature of the resulting solution to fall below room temperature. Which of the following statements is TRUE?a) The products have a lower potential energy than the reactants.b) This type of experiment will provide data to calculate ΔE rxn. c) The reaction is exothermic. d) Energy is leaving the system during the reaction. e) None of the above statements is true. The same reaction in a bomb and coffee-cup calorimeter :a) will give the same value for ΔH rxn because it is the same reaction. b) will give the same value for ΔH rxn because both systems are identical.c) will give the same values because both systems are at a constant pressure. d) will give slightly different values because the coffee-cup calorimeter will do some PV work. e) will give slightly different values because the bomb calorimeter will do some PV work.

A 1.00 g sample of NH 4NO3 is decomposed in a bomb calorimeter. The temperature increases by 6.12°C. What is the molar heat of decomposition for ammonium nitrate?a) -602 kJ•mol-1b) -398 kJ•mol-1c) 7.53 kJ•mol-1d) 164 kJ•mol-1

Which will release more heat as it cools from 50°C to 25°C, 1 kg of water or 1 kg of aluminum? How do you know?

Two solid objects, A and B, are placed in boiling water and allowed to come to the temperature of the water. Each is then lifted out and placed in separate beakers containing 1000 g water at 10.0°C. Object A increases the water temperature by 3.50°C; B increases the water temperature by 2.60°C. (a) Which object has the larger heat capacity?

The specific heat of iron metal is 0.450 J/g-K. How many J of heat are necessary to raise the temperature of a 1.05-kg block of iron from 25.0°C to 88.5°C?

When a 9.55-g sample of solid sodium hydroxide dissolves in 100.0 g of water in a coffee-cup calorimeter (Figure 5.17), the temperature rises from 23.6°C to 47.4°C. Calculate ΔH (in kJ/mol NaOH) for the solution processNaOH(s) → Na+(aq) + OH–(aq)Assume that the specific heat of the solution is the same as that of pure water.

A 2.200-g sample of quinone (C6H4O2) is burned in a bomb calorimeter whose total heat capacity is 7.854 kJ/°C. The temperature of the calorimeter increases from 23.44°C to 30.57°C. What is the heat of combustion per gram of quinone? Per mole of quinone?

A house is designed to have passive solar energy features. Brickwork incorporated into the interior of the house acts as a heat absorber. Each brick weighs approximately 1.8 kg. The specific heat of the brick is 0.85 J/g-K. How many bricks must be incorporated into the interior of the house to provide the same total heat capacity as 1.7 x 103 gal of water?

A coffee-cup calorimeter of the type shown in Figure 5.17 contains 150.0 g of water at 25.1°C. A 121.0-g block of copper metal is heated to 100.4°C by putting it in a beaker of boiling water. The specific heat of Cu(s) is 0.385 J/g•K. The Cu is added to the calorimeter, and after a time the contents of the cup reach a constant temperature of 30.1°C. (a) Determine the amount of heat, in J, lost by the copper block.

A coffee-cup calorimeter of the type shown in Figure 5.17 contains 150.0 g of water at 25.1°C. A 121.0-g block of copper metal is heated to 100.4°C by putting it in a beaker of boiling water. The specific heat of Cu(s) is 0.385 J/g•K. The Cu is added to the calorimeter, and after a time the contents of the cup reach a constant temperature of 30.1°C. (b) Determine the amount of heat gained by the water. The specific heat of water is 4.18 J/g•K.

A coffee-cup calorimeter of the type shown in Figure 5.17 contains 150.0 g of water at 25.1°C. A 121.0-g block of copper metal is heated to 100.4°C by putting it in a beaker of boiling water. The specific heat of Cu(s) is 0.385 J/g•K. The Cu is added to the calorimeter, and after a time the contents of the cup reach a constant temperature of 30.1°C. (c) The difference between your answers for (a) and (b) is due to heat loss through the Styrofoam® cups and the heat necessary to raise the temperature of the inner wall of the apparatus. The heat capacity of the calorimeter is the amount of heat necessary to raise the temperature of the apparatus (the cups and the stopper) by 1 K. Calculate the heat capacity of the calorimeter in J/K.

A coffee-cup calorimeter of the type shown in Figure 5.17 contains 150.0 g of water at 25.1°C. A 121.0-g block of copper metal is heated to 100.4°C by putting it in a beaker of boiling water. The specific heat of Cu(s) is 0.385 J/g•K. The Cu is added to the calorimeter, and after a time the contents of the cup reach a constant temperature of 30.1°C. (d) What would be the final temperature of the system if all the heat lost by the copper block were absorbed by the water in the calorimeter?

A sample of calcium carbonate [CaCO3 (s)] absorbs 45.5 J of heat, upon which the temperature of the sample increases from 21.1oC to 28.5oC. If the specific heat of calcium carbonate is 0.82 J/goC, what is the mass (in grams) of the sample?
A. 0.20
B. 5.0
C. 7.5
D. 410
e. 5.0 x 103

A sample of copper absorbs 43.6 kJ of heat, resulting in a temperature rise of 75°C, determine the mass (in kg) of the copper sample if the specific heat capacity of copper is 0.385 J/g°C.
A) 6.62 kg
B) 1.51 kg
C) 3.64 kg
D) 7.94 kg
E) 1.26 kg

Determine the specific heat capacity of an alloy that requires 59.3 kJ to raise the temperature of 150.0 g alloy from 298 K to 398 K.
A) 4.38 J/g°C
B) 3.95 J/g°C
C) 2.29 J/g°C
D) 1.87 J/g°C
E) 2.53 J/g°C

Calculate the amount of heat (in kJ) necessary to raise the temperature of 47.8 g benzene by 57 K. The specific heat capacity of benzene is 1.05 J/g°C
A) 2.59 kJ
B) 2.86 kJ
C) 3.85 kJ
D) 1.61 kJ
E) 16.6 kJ

Calculate the amount of heat (in kJ) required to raise the temperature of a 79.0 g sample of ethanol from 298 K to 385 K. The specific heat capacity of ethanol is 2.42 J/g°C.
A) 12.9 kJ
B) 57.0 kJ
C) 73.6 kJ
D) 28.4 kJ
E) 16.6 kJ

How much heat, in kJ, is required to raise the temperature of 125 g H 2O from 24.3°C to 64.9°C? The specific heat of water is 4.18 J g-1 °C-1.
A) 21.2 kJ
B) 42.4 kJ
C) 12.7 kJ
D) 33.9 kJ
E) 523 kJ

Calculate the amount of heat necessary to raise the temperature of 12.0 g of water from 15.4°C to 93.0°C.
A) 0.027 J
B) 324 J
C) 389 J
D) 931 J
E) 3,890 J

To raise 232 g of an unknown liquid from 15°C to 60°C, 17.9 kJ of energy are required. What is the specific heat capacity, Cs, of the liquid?
A) 5.41 J/(g ∙ °C)
B) 1.71 J/(g ∙ °C)
C) 3.48 J/(g ∙ °C)
D) 1.10 J/(g ∙ °C)

Given the following balanced equation,2 Na(s) + 2 H2O(l) → 2 NaOH(aq) + H2(g) ∆Hrxn = −368.4 kJwhich statement(s) is/are True? I) The reaction is endothermic.II) The heat associated for reacting 1 mole of H 2O(l) is −184.2 kJ.III) The temperature of the solution increases when the reaction is carried out in a coffee-cup calorimeter. A) I only B) II only C) III only D) both I and II E) both II and III

How much heat is absorbed by a pan made of iron, with a mass of 150 g whose temperature rises from 25.0°C to 100.0°C?

NOTE: YOUR ANSWER SHOULD BE IN KJ/MolWhen 0.535 g of compound X is burned completely in a bomb calorimeter containing 3000 g of water, a temperature rise of 0.363◦C is observed. What is ∆Urxn for the combusion of compound X? The hardware component of the calorimeter has a heat capacity of 3.51 kJ/◦C. The specific heat of water is 4.184 J/g ·◦C, and the MW of X is 56.0 g/mol.1. -426.0662. -610.2963. -1074.214. -898.3525. -988.4256. -760.5147. -589.8938. -487.4289. -927.73110. -875.701

The combustion of toluene has a ∆Erxn of –3.91 x 103 kJ/mol. When 1.55 g of toluene (C7H8) undergoes combustion in a bomb calorimeter, the temperature rises from 23.12 °C to 37.57 °C. Find the heat capacity of the bomb calorimeter.

A chemistry student weighs a rock and finds its mass to be 4.7 g. She then finds that upon absorption of 57.2 J of heat, the temperature of the rock rises from 25 °C to 57 °C. Find the specific heat capacity of the substance composing the rock.

When 1.550 g of liquid hexane (C6H14) undergoes combustion in a bomb calorimeter, the temperature rises from 25.87 to 38.13 °C. Find the ∆Erxn in kJ/mol of hexane. The heat capacity of the bomb calorimeter was found in a different experiment to be 5.73 kJ/°C.

Determine the final temperature when a 25.0 g piece of iron at 85.0°C is placed into 75.0 g of water at 20.0°C. The heat capacity of the iron is 0.450 J/g °C.

Calculate q (in kJ) when 2.00 g of water is heated from 37 °C to 47 °C. The specific heat capacity of water is 4.184 J/g•°C.

Suppose a 50.0 g block of silver (specific heat = 0.2350 J/g.°C.) at 100°C is placed in contact with a 50.0 g block of iron (specific heat = 0.4494 J/g.°C.) at 0°C, and the two blocks are insulated from the rest of the universe. The final temperature of the two blocks.a) will be higher than 50°C.b) will be lower than 50°C.c) will be exactly 50°C.d) is unrelated to the composition of the blocks.e) cannot be predicted.

A 2.78g lead weight, initially at 11.0°C, is submerged in 7.66g of water at 52.3°C in an insulated container. What is the final temperature of both the weight and the water at thermal equilibrium?

Sucrose (table sugar) has the formula C12H22O11 (molar mass = 342.30 g/mol) and a food value of 6.49 kJ/g. Determine the calorimeter constant of the calorimeter in which the combustion of 0.995 g of sucrose raises the temperature by 4.21 degrees celsius (answer must be in kJ/degrees celsius).

If you have 340.0 mL of water at 25.00 °C and add 120.0 mL of water at 95.00 °C, what is the final temperature of the mixture? Use 1.00 g/mL as the density of water.

Which one of the following statements is correct?A) The SI unit of specific heat capacity is calories per gram (cal/g)B) Specific heat capacity is a positive value for liquids and a negative value for solidsC) When heat is transferred from the surrounding to the system q is negativeD) The larger the heat capacity of an object the more thermal energy it can storeE) Heat is transformed from the systems to the surroundings in an endothermic process

5.00 moles of an ideal gas are contained in a cylinder with a constant external pressure of 1.00 atm and at a temperature of 593 K by a movable, frictionless piston. This system is cooled to 504 K.i) Calculate the work done on or by the system. ii) Given that the molar heat capacity (C) of an ideal gas is 20.8 J/mol K, calculate q (J), the heat that flows into or out of the system.

Liquid sodium is being considered as an engine coolant. How many grams of liquid sodium (minimum) are needed to absorb 3.10 MJ of energy (in the form of heat) if the temperature of the sodium is not to increase by more than 10.0. Use Cm = 30.8 J/(K·mol) for Na(l).

What is the molar heat capacity of liquid water? Express your answer using four significant figures.

The specific heat of methanol is 2.5104 J/g.ºC. How many kJ are necessary to raise the temperature of 2.00 L of methanol from 14.0C to 30.0°C? The density of methanol is 0.7915 g/mL.

If a reaction produces 1.506 kJ of heat, which is trapped in 30.0 g of water initially at 26.5°C in a calorimeter like that in Figure 5.12, what is the resulting temperature of the water?

Ethanol (C2H5OH) has been proposed as an alternative fuel. Calculate the standard enthalpy of combustion per gram of liquid ethanol.

A sample of steam with a mass of 0.510 g and at a temperature of 100°C condenses into an insulated container holding 4.50 g of water at 2.0°C.( ΔH°vap = 40.7 kJ/mol, water = 4.18 J/g•°C) Assuming that no heat is lost to the surroundings, what is the final temperature of the mixture?

A hot lump of 106.2 g of an unknown substance initially at 153.2°C is placed in 35.0 mL of water initially at 25.0°C and allowed to reach thermal equilibrium. The final temperature of the system is 53.2°C. What is the identity of the unknown substance? Assume no heat is lost to the surroundings.

A 54.0 g block of an unknown metal is heated in a hot water bath to 100.0°C. When the block is placed in an insulated vessel containing 130.0 g of water at 25.0°C, the final temperature is 28.0°C. Determine the specific heat of the unknown metal. The Cs for water is 4.18 J/g°C.

A student constructs a "coffee cup" calorimeter that contains 83.6 grams of water, at 19.7°C, in a double cup set up with a thermometer and a cork cover. A piece of copper with a mass of 101.7 grams was heated to a certain temperature and placed in the calorimeter. Then the calorimeter was allowed to equilibrate and the thermometer recorded a temperature of 28.3 °C after the equilibration. Determine the temperature to which the copper piece was heated initially. (The specific heat of copper is 0.385 J/g °C and the specific heat of water is 4.184 J/g °C)a. 105.1 °Cb. 85.4 °Cc. 142.0 °Cd. 29.0 °Ce. 48.5 °C

A 0.3423 g sample of pentane, C5H12, was burned in a bomb calorimeter. The temperature of the calorimeter and the 1.000 kg of water contained therein rose from 20.22 °C to 22.82 °C. The heat capacity of the calorimeter is 2.21 kJ/°C. The heat capacity of water = 4.184 J/g•°C. How much heat was given off during combustion of the sample of pentane?1) 8.8 kJ2) -8.8 kJ3) 16.6 J4) 16.6 kJ5) 3.1415 kJ

The specific heat of a certain type of cooking oil is 1.75 J/(g • °C). How much heat energy is needed to raise the temperature of 2.02 kg of this oil from 23 °C to 191 °C?

What is the heat capacity of 155 g of liquid water?

Suppose that 26 g of each of the following substances is initially at 28.0 ºC. What is the final temperature of each substance upon absorbing 2.40 kJ of heat?a) aluminum

When Karl Kaveman adds chilled grog to his new granite mug, he removes 10.9 kJ of energy from the mug. If it has a mass of 625 g and was at 25°C, what is its new temperature? Specific heat capacity of granite = 0.79 J/g • °C. A. 3 °C B. 14 °C C. 22 °C D. 47 °C E. None of these choices is correct.

The specific heat of aluminum is 0.900 J/g • °C. How many joules of heat are absorbed by 15.0 g of Al if it is heated from 20.0°C to 60.0°C?a. 270 Jb. 2.40 Jc. 540 Jd. 812 Je. 0.000117 J

During an experiment, a student adds 1.23 g of CaO to 200.0 mL of 0.500 M HCI. The student observes a temperature increase of 5.10°C. Assuming the solution's final volume is 200.0 mL, the density is 1.00 g/mL, and the heat capacity is 4.184 J/(g.°C), calculate the heat of the reaction (kJ/mol), ΔHrxn.CaO (s) + 2H+ (aq) → Ca2+ (aq) + H2O (l)

A calorimeter was used to measure the heat change when an ionic compound dissolves in water. The final mass of the contents of the calorimeter was 50.73 g and the change in temperature was 2.35 ºC.a) Calculate the heat change of the calorimeter contents, given that its specific heat is 4.10 J/(g ºC).b) Calculate the heat change of the calorimeter, given that its specific heat is 8.5 J/ºC.c) Calculate the heat change of the solution process.

If 125 cal of heat is applied to a 60.0-g piece of copper at 22.0°C, what will the final temperature be? The specific heat of copper is 0.0920 cal/(g • °C). Express your answer with the appropriate units.

Dry ice is solid carbon dioxide. Instead of melting, solid carbon dioxide sublimes according to the following equation: CO2 (s) → CO2 (g). When dry ice is added to warm water, heat from the water causes the dry ice to sublime more quickly. The evaporating carbon dioxide produces a dense fog often used to create special effects. In a simple dry ice fog machine, dry ice is added to warm water in a Styrofoam cooler. The dry ice produces fog until it evaporates away, or until the water gets too cold to sublime the dry ice quickly enough. Suppose that a small Styrofoam cooler holds 15.0 liters of water heated to 83 °C. Part AUse standard enthalpies of formation to calculate the change in enthalpy (kJ) for dry ice sublimation. (The ΔH°f for CO2 (s) is -427.4 kJ/mol). Express your answer using three significant figures. Part BCalculate the mass (grams) of dry ice that should be added to the water so that the dry ice completely sublimes away when the water reaches 18 °C. Assume no heat loss to the surroundings. Express your answer using two significant figures.

An 80.0-gram sample of a gas was heated from 25°C to 225°C. During this process, 346 J of work was done by the system and its internal energy increased by 6565 J. What is the specific heat of the gas?

Suppose you place 0.0300 g of magnesium chips in a coffee-cup calorimeter and then add 100 0 ml. of 1.00 M HCl. The reaction that occurs isMg(s) + 2 HCl(aq) → H2(g) + MgCl2(aq)The temperature of the solution increases from 22.56 °C (295.71 K) to 23.91 °C (297.06 K). What is the enthalpy change (kJ/mol) for the reaction per mole of Mg? Assume a specific heat capacity of the solution is 4.20 J/g • K and the density of the HCI solution is 1.00 g/mL.

In the following experiment, a coffee-cup calorimeter containing 100 mL of H 2O is used. The initial temperature of the calorimeter is 23.0°C. If 9.50 g of CaCl2 is added to the calorimeter, what will be the final temperature of the solution in the calorimeter? The heat of solution ΔHsoln of CaCl2 is -82.8 kJ/mol. Express your answer with the appropriate units.

Typically, water runs through the baseboard copper tubing and, therefore, fresh hot water is constantly running through the piping. However, consider a pipe where water was allowed to sit in the pipe. The hot water cools as it sits in the pipe. What is the temperature change (ΔT) of the water if 202.0 g of water sat in the copper pipe releasing 1760. J of energy to the pipe? The specific heat of water is 4.184 J/(g •°C). Express your answer to four significant figures.

When a solid dissolves in water, heat may be evolved or absorbed. The heat of dissolution (dissolving) can be determined using a coffee cup calorimeter. In the laboratory a general chemistry student finds that when 3.50 g of CuSO4 (s) are dissolved in 115.90 g of water, the temperature of the solution increases from 24.19 to 27.48°C. The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.78 J/°C. Based on the student's observation, calculate the enthalpy of dissolution of CuSO4 (s) in kJ/mol. Assume the specific heat of the solution is equal to the specific heat of water.

Some homes that use baseboard heating use copper tubing. Hot water runs through and heats the copper tubing, which in turn heats aluminum fins. It is actually the aluminum fins that heat the air rising through the fins. How much energy (J) would it take to heat a section of the copper tubing that weighs about 505.0 g, from 13.33°C to 22.38°C? Copper has a specific heat of 0.3850 J/g • °C). Express your answer to four significant figures.

In the laboratory a "coffee cup " calorimeter, or constant pressure calorimeter, is frequently used to determine the specific heat of a solid, or to measure the energy of a solution phase reaction. A student heats 69.07 grams of iron to 97.56°C and then drops it into a cup containing 81.71 grams of water at 22.09°C. She measures the final temperature to be 28.09°C. The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.82 J/°C. Assuming that no heat is lost to the surroundings calculate the specific heat of iron.

A researcher studying the nutritional value of a new candy places a 4.60-gram sample the candy inside a bomb calorimeter and combusts it in excess oxygen. The observed temperature increase is 2.81°C. If the heat capacity of the calorimeter is 38.20 kJ • K-1, how many nutritional Calories are there per gram of the candy?

When a 6.50-g sample of solid sodium hydroxide dissolves in 100.0 g of water in a coffee-cup calorimeter, the temperature rises from 21.6°C to 37.8°C. Calculate ΔH (in kJ/mol NaOH) for the solution process NaOH(s) → Na+ (aq) + OH- (aq) Assume that the specific heat of the solution is the same as that of pure water. Express your answer with the appropriate units.

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