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: Heat capacity vs. Specific Heat Capacity6m
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.
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.
C = m * ∆T
Just multiply both sides by mass * change in temperature.
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.
Concept: Bomb Calorimeter6m
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.
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.
C = ∆T
= - 55 600 J
= - 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: In an experiment a 9.87 carat (1 carat = 0.200g) diamond is heated to 72.25oC and immersed in 22.08 g of water in a calorimeter. If the initial temperature of the water was 31.0oC what is the final temperature of the water? (cdiamond = 0.519) (cwater = 4.184 ).9m
Problem: 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.4m
When dealing with heat it is important to remember that if one compound is gaining heat that means another compound is losing heat.
Problem: 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?5m
The enthalpy or heat of a reaction can be calculated through the use of a coffee cup calorimeter.
Concept: Coffee Cup Calorimeter2m
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: Calculating the Heat of the Solution4m
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: Calculating the enthalpy of the reaction6m
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.
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.
What is the enthalpy change associated with heating 20g of Al from room temperature (23°C) to its melting point at 660°C and then melting it. (Specific heat of Al, Cp = 0.900J/kg °C; ΔH fus = 437 kJ/mole)
Consider a 1 kg block of ice at standard pressure. If it is initially at −5°C and is heated until it is steam at 109°C, how much total heat was added to the sample of water? Use the following thermodynamic values for your calculation:
cice = 2.09 J/g K
cwater = 4.184 J/g K
csteam = 2.03 J/g K
∆Hvap = 2260 J/g
∆Hfus = 334 J/g
1. 3950 kJ
2. 2710 kJ
3. 2620 kJ
4. 3040 kJ
5. 28.7 kJ
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?
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.
When 5.50 g of Ba(s) is added to 100.00 g of water in a container open to the atmosphere, the reaction shown below occurs and the temperature of the resulting solution rises from 22.00°C to 61.16°C. If the specific heat capacity of the solution is 4.18 J/(g∙°C), calculate ΔHrxn for the reaction, as written.
Ba(s) + 2 H2O(l) → Ba(OH)2(aq) + H2(g) ΔHrxn = ?
A) –431 kJ
B) –3.14 kJ
C) +3.14 kJ
D) +431 kJ
E) –17.2 kJ
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.
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
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 °C
b. 85.4 °C
c. 142.0 °C
d. 29.0 °C
e. 48.5 °C
How much energy would be absorbed or released by the H 2O in the process.
30 grams H2O(g) at 100°C → 30 grams H2O(l) at 20°C?
1. 16.2 kcal absorbed
2. 16.2 kcal released
3. There would be no transfer of energy due to the first law of thermodynamics
4. 18.6 kcal released
5. 2.4 kcal released
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
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.
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.
b) Fe and Cu
c) Al and Cu
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.
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?
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
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?
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
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.
The specific heat of liquid ethanol, C2H5OH(l), is 2.46 J/g•°C and the heat of vaporization is 39.3 kJ/mol. The boiling point of ethanol is 78.3°C. The molecular weight of ethanol 46 g/mol. What amount of enthalpy is required to heat 50.0 g of liquid ethanol from 23.0°C to ethanol vapor at 78.3°C?
1) 42.7 kJ
2) 49.5 kJ
3) 179 kJ
4) 1970 kJ
5) 6840 kJ
Assuming the heat released from the combustion of CH 4 can be used with 100% efficiency, how much CH4 (in g) is required to be combusted in the presence of excess oxygen to warm 1.75 kg of water (Cw= 4.184 J/gK) from 25.0 to 98.0 °C? The enthalpy of combustion of CH4 is -802.3 kJ/mol.
A. 56.5 g
B. 46.2 g
C. 28.7 g
D. 10.7 g
E. 7.04 g
A 0.500-L sample of 0.400 M Na2CO3 (aq) is added to 0.500 L of 0.400 M Ba(NO 3)2 (aq) in a calorimeter with a total heat capacity equal to 656 J•K –1 at a constant pressure of one bar. The temperature change is +1.25 K. Use these data to calculate the value of ΔH°rxn for the equation:
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 20.22ºC to 22.82ºC. The heat capacity of the calorimeter is 2.21 kJ/ºC. The heat capacity of water = 4.184 J/gºC. How much heat was given off during combustion of the sample of pentane?
1) 8.8 kJ
2) -8.8 kJ
3) 16.6 J
4) 16.6 kJ
5) 3.1415 kJ
The change in enthalpy (∆H) is a measure of the heat of reaction at
1) Constant temperature.
2) Constant pressure.
3) Constant volume.
4) Constant internal energy.
5) Constant entropy.
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?
What is the minimum mass of ice at 0 °C that must be added to the contents of a can of diet cola (340. mL) to cool the cola to 0 °C from room temperature (20.5 °C)? The heat of fusion for ice is 333 J/g. Assume the cola density is equivalent to water (1 g/mL).
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 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
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.
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 for the metal?
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 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 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.
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
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.
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
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?
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
A bar of hot metal is placed in water in a 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 gk 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)?