A 35-g block of ice at -14°C is dropped into a calorimeter (of negligible heat capacity) containing 400 g of water at 0°C. When the system reaches equilibrium, how much ice is left in the calorimeter? The specific heat of ice is 2090 J(kg K) and the latent heat of fusion of water is 33.5 x 10^4 J/kg.

Answer 1

well, we have .035 kg of ice at -14

we have .4 kg of water at 0

If those really are the numbers,
the answer is that some of the water gets frozen by the ice which is 14 degrees BELOW freezing. At least 35 grams of ice will be left.
How much water can we freeze?

How much heat can we get into the ice between -14 and 0 ?
heat into ice to raise to the melting point
= .035 * 2090 * 14 = 1024 Joules
so how much water at 0 deg can we freeze
with that .1024 * 10^4 Joules
we need 33.5 *10^4 Joules/kg
so we can freeze
.1024/33.5 = .003 kg or 3 grams
so we end up with
35+3 = 38 grams of ice.