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
T_{w} = initial temperature of water
T_{Pb} = initial temperature of lead
T_{f} = final temperature at equilibrium
Since no heat is exchanged with the surroundings:
heat gained by the water = heat lost by the lead
mc_{w}(T_{f}  T_{w}) = mc_{Pb}(T_{Pb}  T_{f}) (1)
We can perform the experiment in 2 ways:
1. Solving for c_{Pb} in the heat balance equation (1) gives:
c_{Pb}  = c_{w}  ( 

) 
Now use c_{w} = 4186 J/(kg °C) for the specific heat of water and
the measured value of T_{f} to compute c_{Pb}.
The value for c_{Pb} is 130 J/(kg °C).
2. Alternatively, we use c_{w} = 4186 J/(kg °C) and c_{Pb}
= 130 J/(kg °C) and solve for T_{f} in the heat balance equation
(1). This gives:
T_{f} =  ( 

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