Specific Heat Capacity - Heat Absorbed - Physics - Thermodynamics

Hi there,

This question has been really bugging me for a while now. It is from a past exam paper and I am unsure why I cannot get the correct answer. The question is below:

An explorer takes 200 g of ice from an environment where the temperature is -40 degrees C, and places it in a kettle inside his mobile laboratory. The ice melts and he heats the water to +20 degrees C.
What is the total amount of heat absorbed by the explorer's sample of H20?

A. 543 kJ
B. 109 J
C. 100 kJ
D. 41.9 kJ
E. 83.5 J

Specific Heat Capacity of Ice: 2.09 Kj/Kg/C
Specific Heat Capacity of Water: 4.19 Kj/Kg/C

My attempt:
Change in temperature from -40 -> 0 = 40 degrees
40 degrees * 2.09 * .200 kg = 16.72 kJ

+

Change in temperature from 0 -> 20 degrees = 20 degrees
20 degrees * 4.19 * .200 kg = 16.76 kJ

For a total of 16.72 + 16.76 =33.48 kJ which is not one of the options.

Can anyone spot where I am going wrong please?

your attempt is right, but you need to consider the latent heat of fusion,

for ice, it is around 334 kJ/kG

for 200g = 0.2*334 = 66.8


add it to your answer 33.48

ie, 33.48+66.8 = 100.2kJ

means option C is right

You must also take into account the ice melting. It goes from a solid to a liquid in water. So you need to also include the equation for latent heat of fusion. You also need another equation to account for the specific heat of the liquid which melted from the ice and mixed with the water inside the kettle. So you really need 4 equations to this problem.