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Question: When a piece of metal weighing \( 48.3\,g \) at \( 10.7^\circ C \) is immersed in a current of steam...

When a piece of metal weighing 48.3 g48.3\,g at 10.7∘C10.7^\circ C is immersed in a current of steam at 100∘C100^\circ C , 0.762 g0.762\,g of the steam condenses on it. Calculate the specific heat of the metal.
A) 0.035 cal/gm∘C0.035\,cal/gm^\circ C
B) 0.095 cal/gm∘C0.095\,cal/gm^\circ C
C) 0.125 cal/gm∘C0.125\,cal/gm^\circ C
D) 0.145 cal/gm∘C0.145\,cal/gm^\circ C

Explanation

Solution

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=mLQ = mL where QQ is the amount of energy needed to convert the state of matter of a substance of mass mm and latent heat capacity LL
Q=mcΔTQ = mc\Delta T where QQ is the amount of energy needed to change the temperature of a substance of mass mm and specific heat capacity cc by temperature ΔT\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∘C100^\circ C . We also know that in the act of doing so, 0.762 g0.762\,g of the steam condenses on the surface of the metal. So, the energy required to convert 0.762 g0.762\,g of steam into water will be equal to
Q=0.762×540Q = 0.762 \times 540
⇒Q=411.48 cal\Rightarrow Q = 411.48\,cal
This much heat will be responsible to raise the temperature of the metal block from 10.7∘C10.7^\circ C to 100∘C100^\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=mcΔTQ = mc\Delta T as
411.48=48.3×c×(100−10.7)411.48 = 48.3 \times c \times (100 - 10.7)
Which give us
c=0.095 cal/gm∘Cc = 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∘C100^\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∘C100^\circ C as both the metal block and the steam vapour will have the same temperature.