A 300 gram lead ball with a temperature of 80°C is placed in 300 grams of water at a temperature of 20°C. When the system reaches equilibrium what is the equilibrium temperature? Assume no energy is exchanged with the surroundings.
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. Solving for cPb in the heat balance equation (1) gives:
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:
The final temperature is a weighted average
of the two starting temperatures, with the weighting biased to the
material with the larger specific heat.