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.

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

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. 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 c_Pb in the heat balance equation (1) gives:

cPb

= cw

(

Tf - Tw TPb - Tf

)

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:

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.