Question

I take 1.0 kg of ice and dump it into 1.0 kg of water and, when equilibrium is reached, I have 2.0 kg of ice at 0°C. The water was originally at 0°C.
The specific heat of water = 1.00 kcal/kg⋅°C, the specific heat of ice = 0.50 kcal/kg⋅°C, and the latent heat of fusion of water is 80 kcal/kg.

The original temperature of the ice was:
a. one or two degrees below 0°C.b. −80°C.c. −160°C.d. The whole experiment is impossible.

Answer 1

Answer:

.c. −160°C

Explanation:

In the whole process one kg of water at  0°C loses heat to form one kg of ice and heat lost by them is taken up by ice at −160°C . Now see whether heat lost is equal to heat gained or not.

heat lost by 1 kg of water at  0°C

= mass x latent heat

= 1 x 80000 cals

= 80000 cals

heat gained by ice at −160°C to form ice at  0°C

= mass x specific heat of ice x rise in temperature

= 1 x .5 x 1000 x 160

= 80000 cals

so , heat lost = heat gained.