Heat is energy transferred between a system and its surroundings because of a temperature difference between them.

The specific heat of a material is the amount of heat required to raise 1 kg of the material by 1° C. The symbol for specific heat is c.

Heat lost or gained by an object is given by:

Q = mcΔT

Changes of state occur at particular temperatures, so the heat associated with the process is given by:

freezing or melting: Q = mL_f

where L_f is the latent heat of fusion

Boiling or condensing: Q = mL_v

where L_v is the latent heat of vaporization

For water the values are:

L_v = 2256 kJ/kg

L_f = 333 kJ/kg

c = 4.186 kJ/(kg °C)

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)