Question

If a vessel of thermal capacity $ 30 $ calories contains $ 170g $ of water at $ 30^\circ C $ . If on dropping a solid at $ 93^\circ C $ into the calorimeter, then the temperature of mixture becomes $ 33^\circ C $ , find the thermal capacity of the solid.

Answer 1

Hint : In a calorimeter, the object is dropped into water, on doing so it transfers the heat energy present in it to the water, this causes the overall temperature of the water to rise, thus the heat lost by the hot object is equal to heat gained by the cold object. Here the dropped solid is the hot object while water is the cold object.

Formula used:
 $ \Rightarrow \eta \times m = \omega $
Where $ \eta $ is the specific heat capacity of a substance, $ \omega $ is the heat capacity of the body, $ m $ is the mass of the body.

Complete step by step answer
When the solid is dropped into water, all of its heat is transferred to water if it is assumed that heat does not dissipate into the surroundings.
Therefore it can be written as,
Heat lost by the solid = Heat gained by the water
It is given that initial temperature of the hot body $ {T_i} = 93 $
Final temperature of the hot body $ {T_f} = 33 $
Let the change in temperature of the hot body be $ \Delta T $ . Then,
 $ \Rightarrow \Delta T = {T_f} - {T_i} $ of the solid.
 $ \Rightarrow \Delta {T_s} = 33 - 93 = 60 $
Let the thermal capacity of the solid be $ {\omega _s} $
Then the heat lost by the solid $ {H_L} $ is given by-
 $ \Rightarrow Q = {\omega _s}\Delta {T_s} $
 $ \Rightarrow Q = {\omega _s} \times 60 $
For the water and the vessel (or the cold body),
The initial temperature $ {T_i} = 30 $
The final temperature $ {T_f} = 33 $
The change in temperature of the water in the vessel is,
 $ \Rightarrow \Delta T = {T_f} - {T_i} $ of the water.
 $ \Rightarrow \Delta {T_w} = 33 - 30 = 3 $
The heat gained by the system of water and the vessel can be calculated by adding the individual heat gains of the water and the vessel.
For the vessel, heat gained is,
 $ \Rightarrow {Q_v} = {\omega _v}\Delta {T_w} $
Where, $ {\omega _v} $ is the heat capacity of the vessel equal to $ 30cal $ .
And heat gained by the water is,
 $ \Rightarrow {Q_w} = {\eta _w} \times {m_w} \times \Delta T $
 $ {\eta _w} $ is the specific heat capacity of the water which is equal to $ 1cal/g^\circ C $ .
 $ {m_w} $ is the mass of water, which is given as $ 170g $
Therefore,
The heat gained by water and vessel is given by-
 $ \Rightarrow {Q_v} + {Q_w} = \Delta {T_w}({\omega _v} + {m_w}{\eta _w}) $
On putting the values,
 $ \Rightarrow {Q_v} + {Q_w} = 3(30 + 170 \times 1) $
 $ \Rightarrow {Q_v} + {Q_w} = 600 $
Since this is equal to the heat lost by the solid,
 $ \Rightarrow {Q_v} + {Q_w} = {Q_s} $
 $ \Rightarrow 600 = 60{\omega _s} $
 $ \therefore {\omega _s} = 10 $
The thermal capacity of the solid is $ 10cal/^\circ C $ .

Note
The terms thermal capacity and heat capacity have the same meaning. But the terms heat capacity and specific heat capacity are different. Heat capacity is the heat required to raise the temperature of a body by $ 1^\circ C $ , while the specific heat capacity is the heat required to raise the temperature of $ 1 $ gram of a body by $ 1^\circ C $ .