Three lead fishing weights, each with a mass of 1.00x10^2 g and at a temperature of 100.0oc, are placed in 2.00x10^2 g of water at 35.0oc. the final temperature of the mixture is 45.0oc. what is the specific heat of the lead in the weights?

The lead and the water reach thermal equilibrium when the net flow of heat from the lead to the water becomes zero, so when the heat released by the lead  is equal to the heat absorbed by the water :
(1) 
where the negative sign means the heat of the lead is released, not absorbed.

The equation that gives the heat absorbed/released is
(2)
where
m is the mass of the substance
C is its specific heat capacity
is the variation of temperature.

If we rewrite (1) by using (2), we get
(3)
where the label L refers to the lead and the label W refers to the water.

The total mass of the lead is:

while its change of temperature is


The mass of the water is

its specific heat capacity is

while its change of temperature is


If we put all this data inside (3), we can calculate the specific heat capacity of lead:

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