Question

When a piece of metal weighing $ 48.3\,g $ at $ 10.7^\circ C $ is immersed in a current of steam at $ 100^\circ C $ , $ 0.762\,g $ of the steam condenses on it. Calculate the specific heat of the metal.
A) $ 0.035\,cal/gm^\circ C $
B) $ 0.095\,cal/gm^\circ C $
C) $ 0.125\,cal/gm^\circ C $
D) $ 0.145\,cal/gm^\circ C $

Answer 1

Hint: We will use the concepts of latent heat and specific heat capacity to determine the specific heat of the metal. Since the metal is immersed in a current of steam, its final temperature will be equal to the temperature of the steam.

Formula used : In this solution, we will use the following formula:
 $ Q = mL $ where $ Q $ is the amount of energy needed to convert the state of matter of a substance of mass $ m $ and latent heat capacity $ L $
 $ Q = mc\Delta T $ where $ Q $ is the amount of energy needed to change the temperature of a substance of mass $ m $ and specific heat capacity $ c $ by temperature $ \Delta T $

Complete step by step answer:
We’ve been given that a piece of metal is immersed in a current of steam whose temperature is $ 100^\circ C $ . We also know that in the act of doing so, $ 0.762\,g $ of the steam condenses on the surface of the metal. So, the energy required to convert $ 0.762\,g $ of steam into water will be equal to
 $ Q = 0.762 \times 540 $
 $ \Rightarrow Q = 411.48\,cal $
This much heat will be responsible to raise the temperature of the metal block from $ 10.7^\circ C $ to $ 100^\circ C $ . This is because to establish a temperature equilibrium, the temperature of the metal block will rise to the temperature of the water vapour.
So, we can write $ Q = mc\Delta T $ as
 $ 411.48 = 48.3 \times c \times (100 - 10.7) $
Which give us
$ c = 0.095\,cal/gm^\circ C $ which corresponds to option (B).

Note:
We must notice that the final temperature of the metal block will be $ 100^\circ C $ since the constant stream of water vapour will increase its temperature. The condensation of the vapour on the metal block will also stop when the metal block reaches $ 100^\circ C $ as both the metal block and the steam vapour will have the same temperature.