Question

A 0.50 kg lump of ice at -18 ºC is dropped into a well-insulated beaker with 1.0 kg of water at 42
ºC in it. Calculate what the beaker contains once thermal equilibrium has been reached?

(The specific heat capacities for ice and water are 2220 J kg-1 K-1 and 4190 J kg-1 K-1 respectively. The latent heat of fusion for water is 333 x 103
J kg-1)

Answer 1

Answer on Question #42234, Physics, Molecular Physics — Thermodynamics — for completion

Question:A 0.50 kg lump of ice at -18 ?C is dropped into a well-insulated beaker with 1.0 kg of water at 42?C in it. Calculate what the beaker contains once thermal equilibrium has been reached? (The specific heat capacities for ice and water are 2220 J kg-1 K-1 and 4190 J kg-1 K-1 respectively. The latent heat of fusion for water is 333 x 103 J kg-1)

Solution

There will be less ice, and more water, both at 0^{∘} C. It easy to check. Indeed, when cooling to 0^{∘} water will give amount of heat of

$Q_{1}=c_{w}m_{w}t_{w}=4190\cdot 1\cdot 42=175980J$

Heat required for ice to get to 0^{∘} C is

$Q_{2}=c_{i}m_{i}t_{i}=2220\cdot 0.5\cdot 18=19980J$

Hence, for melting ice it will be left

$Q_{2}-Q_{1}=175980-19980=156000J$

And with this amount of heat you can melt

$\frac{156000}{333000}\approx 0.47\,kg$

this much ice. So, it will be left 0.03 kg of ice and 1.47 kg of water at 0^{∘} C