A 2.0-102 g piece of copper at a temperature of 450 K and a 1.0 . 102 g piece of aluminium at a temperature of 2.0.102 K are dropped into an insulated bucket containing 5.0 .102 g of water at 280 K. What is the equilibrium temperature of the mixture?
THINK:
The copper piece will give up heat and the water and the aluminum piece will absorb heat until all are at the same temperature. Since heat always transfer from the high temperature body to the low temperature body.
SKETCH:
The figure shows the problem situation before and after the copper and aluminum pieces are added to the water.
RESEARCH:
Let the equilibrium temperature of the system is’
’.
The heat lost by the copper piece is given by the equation

Here
and
are the mass, specific heat and the temperature of the copper respectively.
The heat gained by the water and aluminum is given by the equation

Here
and
are the mass, specific heat and the temperature of the aluminum respectively and
, and
are the mass, specific heat and the temperature of the water respectively.
The sum of the heat lost by the copper piece and the gained by the water and aluminum piece is zero. So we can write

SIMPLIFY:
By solving the above equation, we get the equilibrium temperature of the mixture

...... (1)
GIVEN DATA:
The mass of the copper piece,
The specific heat of the copper,
The initial temperature of the copper piece, 
The mass of the water,

The specific heat of the water,
The initial temperature of the water,
The mass of the aluminum piece,

The specific heat of the aluminum,
The initial temperature of the copper, 
CALCULATE:
On substituting the numerical values in equation (1), we get the equilibrium temperature of the mixture
Therefore, the equilibrium temperature of the mixture is
.