Question

24. A 0.0500-kg ice cube at −30.0ºC is placed in 0.400 kg
of 35.0ºC water in a very well-insulated container. What is
the final temperature?

Answer 1

use! Heat capacity of water is ABOUT 4.19 kJ/(kg K),

and that of ice is ABOUT 2.1 kJ/(kgK),

but these vary over the range of temperatures in the problem, while your book and teacher are undoubtedly expecting you to use some standard value, which you were probably given.

If I use the values above, I get that it take 3.15 kJ to bring the ice up to 0C without melting it;

and the water would give off 66.0 kJ in cooling down to 0C without freezing.

It would take 333.6 kJ/kg to melt the ice,

so that would be (0.0500kg)(334 kJ/kg) = 16.68 kJ to melt the ice completely.

Apparently the ice will melt completely, absorbing a total of 19.83 kJ.

That 19.83 kJ would cool the water only by

(19.83 kJ)/[(0.400 kg)(4.19 kJ/(kg K))] = 11.83K,

so you can conceive of this as a mixture of 0.400 kg of water at 24.48C and 0.05 kg of water at 0C.

The water at 0C will warm up by 9K for every 1K that the water at 24.48C cools down.

So the final temperature is about 20.6C.