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 sure

We'll actually do the experiment to see. Let's stick with 300 grams for the two masses, and use the following:

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:

ΣQ = 0

mc_w(T_f - T_w) + mc_Pb(T_f - T_Pb) = 0

The masses cancel because they're equal in this case. We'll measure all the temperatures, so we can determine the specific heat of lead if we know that the specific heat of water is:

c_w = 4186 J/(kg °C)

Our equation above becomes:

c_Pb(T_f - T_Pb) = -c_w(T_f - T_w)

cPb

= cw *

(

Tf - Tw TPb - Tf

)

The accepted value for c_Pb is 130 J/(kg °C)