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. q1 which is the heat in raising the temperature of 12.0 g of ice from –5.0˚C to 0˚C
Step 2. q2 which is the heat in melting 12.0 g of ice at 0˚C
Step 3. q3 which is the heat in raising the temperature of 12.0 g of water from 0˚C to 0.5˚C
Step 4. Calculate Qtotal.
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:
ΔHfus= 6.02 kJ/mol, cliquid = 4.18 J/g °C, csolid = 2.09 J/g °C
A. –2.2 x 103 J
B. 2.1 x 102 J
C. –2.1 x 102 J
D. 4.2 x 103 J
E. 2.2 x 103 J