Problem solving strategies:
calorimetry and specific heat

The problems we will be working on typically involve a sample of hot metal put into a container of water, where you are asked to solve for the equilibrium temperature. It is useful to estimate what the answer will be based on the specific heats of the materials involved. Because water has such a high specific heat compared to other things, its temperature probably will not change as much as the other parts of the system during heat exchange. As a result, the equilibrium temperature usually turns out to be close to the starting temperature of the water.

Our example: suppose we have a 100g aluminum cup with 80g of liquid water in it, both at a temperature of 0oC; to these we add:

  1. 30g of ice at initial temperature -40o.
  2. 150g of hot aluminum ingots at initial temperature 99o.

What will the equilibrium temperature be? Before we can answer, we have to figure out what will happen to the ice. If all of the ice melts, it will be possible for the system to have an equilibrium temperature higher than 0o. If only some of the ice melts, the equilibrium temperature will automatically be 0o.

First, let's assume that just enough heat is supplied to the ice to melt it. First heat is added to raise the temperature of the ice to 0o, then heat is added to melt the ice at constant temperature:

Q = mcDT + mLf
= (30g)(0.5cal/goC)[0oC-(-40oC)] + (30g)(79.7cal/g)
= 2991 cal.

Where might the heat come from to accomplish this ? The only possibility is the hot ingots; so let's figure out the maximum amount of heat that could be extracted from the ingots (using the lowest possible equilibrium temperature):

Q = mcDT = (150g)(0.22cal/goC)(0oC-99oC)
= -3267 cal.

From these calculations, we conclude that excess heat is available from the ingots after all of the ice melts, and so the temperature of the system will rise above 0o. Now we have to set up the equation of heat exchange with this knowledge. The input side includes heat to warm the ice, heat to melt the ice, heat to warm all of the water (including what came from ice), and heat to warm the aluminum cup. The output side is just the heat given off by the hot ingots. Input heat equals output heat:

[(30)(0.5)(40) + (30)(79.7) + (110)(1.0)(Teq-0o) + (100)(0.22)(Teq-0o)]
= [(150)(0.22)(99o-Teq)]

[600 + 2391 + 110T - 0 + 22T] = [3267 - 33T]
165T = 276
Teq = 1.7o.

As expected, the equilibrium temperature is not far above the starting temperature of the water.

Back to Topics Outline