We’re being asked to calculate the **amount of heat** needed to convert **12.0 g of ice at -5.0˚C to liquid water at 0.5°C**. There are three heats involved in this problem:

**Step 1.** **q _{1}** which is the heat in raising the temperature of 12.0 g of ice from

**Step 2.** **q _{2}** which is the heat in melting 12.0 g of ice at

**Step 3. ****q _{3}** which is the heat in raising the temperature of 12.0 g of water from

We need to solve for each heat individually then add them together to get the final answer.

**Step 1****. The temperature changes from –5.0˚C to 0˚C****. **

Because of this, the equation we’ll use is:

From the data below, calculate the total heat (in J) needed to convert 12.0 g of ice at -5.0 °C to liquid water at 0.5 °C:

ΔH_{fus}= 6.02 kJ/mol, c_{liquid} = 4.18 J/g °C, c_{solid} = 2.09 J/g °C

A. –2.2 x 10^{3} J

B. 2.1 x 10^{2} J

C. –2.1 x 10^{2} J

D. 4.2 x 10^{3} J

E. 2.2 x 10^{3} J

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Heating and Cooling Curves concept. You can view video lessons to learn Heating and Cooling Curves. Or if you need more Heating and Cooling Curves practice, you can also practice Heating and Cooling Curves practice problems.