Example Problem

A 300 gram lead ball with a temperature of 80C is placed in 300 grams of water at a temperature of 20C. When the system reaches equilibrium what is the equilibrium temperature? Assume no energy is exchanged with the surroundings.

  1. a little less than 80C
  2. around 50C
  3. a little more than 20C
  4. not enough information is given















Let's do the calculation to see. Let

Tw = initial temperature of water
TPb = initial temperature of lead
Tf = final temperature at equilibrium

Since no heat is exchanged with the surroundings:

heat gained by the water = heat lost by the lead

mcw(Tf - Tw) = mcPb(TPb - Tf) (1)

We can perform the experiment in 2 ways:

  1. From the final temperature and the specific heat of water, determine the specific heat of lead.
  2. Compute the final temperature given the specific heats of water and lead.


1. Solving for cPb in the heat balance equation (1) gives:

cPb = cw (
Tf - Tw
TPb - Tf
)

Now use cw = 4186 J/(kg C) for the specific heat of water and the measured value of Tf to compute cPb.


The value for cPb is 130 J/(kg C).

2. Alternatively, we use cw = 4186 J/(kg C) and cPb = 130 J/(kg C) and solve for Tf in the heat balance equation (1). This gives:

Tf = (
cPb TPb + cw Tw
cPb + cw
)

The final temperature is a weighted average of the two starting temperatures, with the weighting biased to the material with the larger specific heat.