We’re being asked to calculate the amount of energy required to convert 70.6 g of ice at 0.00˚C to water at 75˚C.
There are three heats involved in this problem:
1. q1 which is the heat in melting 70.6 g ice at 0.0°C.
2. q2 which is the heat in raising the temperature of 70.6 g of water from 0.00˚C to 75˚C.
We need to solve for each heat individually then add them together to get the final answer.
How much energy is needed to convert 70.6 grams of ice at 0.00°C to water at 75.0°C?
specific heat (ice) = 2.10 J/g°C
specific heat (water) = 4.18 J/g°C
heat of fusion = 333 J/g
heat of vaporization = 2258 J/g
A. 45.6 kJ
B. 2.53 kJ
C. 182 kJ
D. 22.1 kJ
E. 34.6 kJ