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