Question

You drop a large ice cube of mass mc = 200g at an initial temperature of Tc = -4 degreesC into mw = 200g of water at Tw = 14 degreesC.

(a) Find the heat lost by the water if it is cooled to exactly T = 0 degreesC.

(b) Find the heat required to raise the temperature of the ice cube to it's melting point. Assume cice = 2100J=kgCdegrees

(c) With the additional energy, find how much ice is to be melted to bring the water down to T = 0 degreesC.

Answer 1

(a)

The heat Q lost by the water as it cools is given by , where m is the mass of the water, c is the specific heat of water, and is the change in temperature.

Plug , , , and into to find the heat lost by the water.

The heat lost by the water is 11732 joules.

(b)

To find the heat required to raise the ice cube up to its melting point, plug , , , and into .

Notice that this is much smaller than the energy required to cool the liquid water down. The ice cube was already close to its melting point while the liquid water was much further from 0 degrees Celsius.

(c)

The heat lost by the liquid water is gained by the ice so that it melts (and then reaches an equilibrium temperature of 0 degrees Celsius). The heat in the system is the heat needed to raise the ice to its melting temperature, the heat needed for the phase change of the ice into liquid, and then the heat lost by the original liquid water. The heat required for a phase change is given by where L is the heat of fusion. The equilibrium heat equation is thus:

Simplify and rearrange to solve for the mass of the ice:

Now plug in , , and .

The amount of ice that would need to be added to cool the water to that temperature while warming and then melting the ice is 0.0343 kg or 34.3 g.