How much heat (kJ) must be removed in order to change 2.78 moles of water at 49.0 oC to ice at -24.0

Question

How much heat (kJ) must be removed in order to change 2.78 moles of water at 49.0 oC to ice at -24.0 oC ?


The specific heat capacity of solid water is 2.03 J g-1 oC -1

Answer 1

You're going to need more information to solve the problem.

Overall, you're going to have to think about this problem in 3 steps.

1. How much heat must be removed to cool 2.78 moles of water from 49oC to 0oC?
2. How much heat must be removed to convert 2.78 moles of water into ice at 0oC?
3. How much heat must be removed to cool 2.78 moles of ice from 0oC to -24oC?

For the first and the last, just use q=mcDT where m is the mass, c is the specific heat (heat capacity) and DT is (T2-T1) (final temperature - initial temperature). For the second one, you'll need to have the heat of fusion of water, which I'm sure you can find in another problem, or in a table in your text.

Answer 2

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