Question

512 g of an unknown metal at a temperature of 15^oC is dropped into a 100 g aluminum container holding 325 g of water at 98^oC. A short time later, the container of water and metal stabilizes at a new temperature of 78^oC. Identify the metal.  

Answer 1

Here we will apply the conservation of energy. It means heat gained by the metal = Heat lost by the water.

Heat gained by the metal = heat loss by water + calorie meter

Mass of metal $(M_1) = 512g$

Mass of water $(M_2) = 325g$

Initial temperature of the metal $(T_1) = 15°C$

Initial temperature of water $(T_2) = 98°C$

Final temperature of the mixture $(T_3) = 78°C$

Specific heat capacity of metal $(C_1) = ?$

Specific heat capacity of water $(C_2) = 4.184J/g°C$

Heat loss by water = heat gain by the metal

$M_2C_2(T_2 - T_3) = M_1C_1(T_3 - T_1)$

$\Rightarrow 325 × 4.184 × (98 - 78) = 512 × C_1 × (78 - 15)$

$\Rightarrow 1359.8 × 20 = 512C_1 × 63$

$\Rightarrow 27196 = 32256C_1$

$\Rightarrow C_1 = 27196 / 32256$

$\Rightarrow C_1 = 0.843J/g°C$

The specific heat capacity of the metal is $0.843J/g°C$