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
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.