A sphere of `0.047` kg aluminium is placed for sufficient time in a vessel containing boiling water , so that the sphere is at `100 ^(@) C` . It is then immediately transferred to `0.14` kg copper calorimeter containing `0.25` kg water at `20^(@) C` . The temperature of water rises and attains a steady state at `23 ^(@) C`. Calculate the specific heat capacity of aluminium.

Explanation

In solving this example, we shall use the fact that at a steady state, heat given by an aluminium sphere will be equal to the heat absorbed by the water and calorimeter.
Mass of aluminium sphere `(m_(1)) = 0.047 ` kg Initial temperature of aluminium sphere `= 100 ^(@) C`
Final temperature `= 23^(@) C`
Change in temperature `(Delta T) = (100 ^(@) C - 23 ^(@) C) = 77 ^(@) C`
Let specific heat capacity of aluminium be `s_(Al)`.
The amount of heat lost by the aluminium sphere = `m_(1) s_(Al) Delta T = 0.047 kg xx s_(Al) xx 77 ^(@) C`
Mass of water `(m_(2)) = 0.25 kg`
Mass of calorimeter `(m_(3)) = 0.14 kg`
Initial temperature of water and calorimeter = `20^(@) C`
Final temperature of the mixture ` = 23 ^(@) C`
Change in temperature `(Delta T_(2)) = 23 ^(@) C - 20 ^(@) C = 3 ^(@) C`
Specific heat capacity of copper calorimeter ` = 0.386 xx 10^(3) J kg^(-1) K^(-1)`
Specific heat capacity of copper calorimeter
` = 0.386 xx 10^(3) J kg^(-1) K^(-1)`
The amount of heat gained by water and calorimeter ` = m_(2) s_(w) Delta T_(2) + m_(3) s_(cu) Delta T_(2)`
`= (m_(2) s_(w) + m_(3) s_(su)) (Delta T_(2))`
` = (0.25 kg xx 4.18 xx 10^(3) J kg ^(-1) K^(-1) + 0.14 kg xx 0.386 xx 10^(3) J kg^(-1) K^(-1) (23 ^(@) C - 20 ^(@) C)`
In the steady state heat lost by the aluminium sphere = heat gained by water `+ ` heat gained by calorimeter.
So, `0.047 kg xx s_(Al) xx 77^(@) C`
` = (0.25 kg xx 4.18 xx 10^(3) J kg^(-1) K^(-1) + 0.14 kg xx 0.386 xx 10^(3) J kg^(-1) K^(-1)) ( 3^(@) C)`
` s_(Al) = 0.911 kJ kg^(-1) K^(-1)`