Question

A 10 kg block of ice has a temperature of -13°C. The pressure is one atmosphere. The block absorbs 4.10 106 J of heat. What is the final temperature of the liquid water?

Answer 1

Lets do it for one kg and say it absorbs 4.1 * 10^5 Joules to make the numbers easier.

first raise temp of ice from -13 to 0

Joules = 2000 * 13 = 26,000 = .26*10^5
so we have 4.1 - .26 = 3.84*10^5 Joules left and 1 kg ice at zero C

Now melt the ice
Joules = 3.34 10^5 Joules/kg heat of fusion
so we have (3.84 -3.34) = .5*10^5 Joules left and water at zero deg C

now heat the water
Joules = 5*10^4 = .4190*10^4 * (T-0)
T = 11.9 deg C

Answer 2

How much heat is required to bring the block of ice up to 0C?
How much heat is required to melt the ice at 0C?
If there is heat left over, then final temp can be calculated by

Heatleftover=10kg*specificheatwater(Tf-0)

solve for Tf.

The statement "the block absorbs 4.10106 J of heat" makes no sense to me.

Answer 3

calories given off when 87g {\rm g} of water cools from 47 ∘ C ^\circ C to 24 ∘ C