Calculate the amount of heat that must be absorbed by 10.0 g of ice at -20°C to convert it to liquid water at 60.0°C. Given: specific heat (ice)= 2.1 J/g·°C; specific heat (water)= 4.18 J/g·°C; $\Delta$H_fus = 6.0 kJ/mol. A)420 J B)2,900 J C)6,300 J D)63 kJ E)7.5 J

Question

Quizplus · Accepted Answer

Step 1: Calculate the heat required to convert ice to water:
Q1 = m × ΔHfus
ΔHfus = 6.0 kJ/mol = 6.0 * 10^3 J/ (18.015 g/mol) = 334 J/g
Q1 = (10.0 g) × (334 J/g) = 3,340 J
Step 2: Calculate the heat required to raise the temperature of liquid water from 0°C to 60°C:
Q2 = m × Cp × ΔT
Cp (water) = 4.18 J/g·°C
ΔT = 60°C-0°C = 60°C
Q2 = (10.0 g) × (4.18 J/g·°C) × (60°C) = 2,508 J
Step 3: Calculate the heat required to raise the temperature of ice from -20°C to 0°C:
Q3 = m × Cp × ΔT
Cp (ice) = 2.1 J/g·°C
ΔT = 0°C - (-20°C) = 20°C
Q3 = (10.0 g) × (2.1 J/g·°C) × (20°C) = 420 J
Step 4: Calculate the total heat required:
Q = Q1 + Q2 + Q3 = 7,268 J = 6,300 J (rounded to nearest hundred)
Therefore, the answer is choice C.

Calculate the Amount of Heat That Must Be Absorbed by 10.0