Question

Two identical calorimeters $A$ and $B$ contain equal quantity of water at ${20}^{\circ }C$. A $5gm$ piece of metal $X$ of specific heat $0.2cal{g}^{-1}(C^{\circ }{)}^{-1}$ is dropped into $A$ and a $5gm$ piece of metal $Y$ into $B$. The equilibrium temperature in $A$ is ${22}^{\circ }C$ and in $B,{23}^{\circ }C$. The initial temperature of both the metals is ${40}^{\circ }C$. Find the specific heat of metal $Y$ in $cal{g}^{-1}(C^{\circ }{)}^{-1}$.

• A. $\frac{27}{85}$
• B. $\frac{37}{85}$
• C. $\frac{27}{75}$
• D. $\frac{37}{75}$

Answer 1

The correct option is A $\frac{27}{85}$

Principle of calorimeter:-

Heat lost to metal = Heat gained by water

For metal X,

$m_w\times (1cal/g-C)\times (22-20)=5\times \left(0.2\right)\times \left(18\right)\Rightarrow m_w=9gms$

For metal Y,

$9\times \left(1\right)\times (23-20)=5\times C_Y\times (40-23)C_Y=\frac{27}{85}cal/g-C$

Hence,option A is correct.