Question

Two identical calorimeters A and B contain an equal quantity of water at 20^oC. A 5g piece of metal X of specific heat 0.2 cal g^-1o C^-1 is dropped into A and a 5g piece of metal Y into B. The equilibrium temperature in A is 22^o C and in B is 23^o C. The initial temperature of both the metals is 40^o C. Find the specific heat of metal Y in cal g^-1oC^-1.

• A. $\frac{27}{85}\raisebox{1ex}{$cal$}\!\left/ \!\raisebox{-1ex}{$g°C$}\right.$
• B. $26\raisebox{1ex}{$cal$}\!\left/ \!\raisebox{-1ex}{$g°C$}\right.$
• C. $\frac{85}{27}\raisebox{1ex}{$cal$}\!\left/ \!\raisebox{-1ex}{$g°C$}\right.$
• D. $27\raisebox{1ex}{$cal$}\!\left/ \!\raisebox{-1ex}{$g°C$}\right.$

Answer 1

The correct option is A

$\frac{27}{85}\raisebox{1ex}{$cal$}\!\left/ \!\raisebox{-1ex}{$g°C$}\right.$

Step 1: Given data

The temperature of water in calorimeters A and B, $T_w={20}^{o}C$

Mass of metals X and Y, $m=5g$

Specific heat of metal X, $s_x=0.2cal{g}^{-1o}{C}^{-1}$

The initial temperature of both the metals, $T_1={40}^{o}C$

Equilibrium temperature in calorimeter A, $T_A={22}^{o}C$

Equilibrium temperature in calorimeter A, $T_B={22}^{o}C$

Step 2: Finding the mass of water

According to the principle of calorimetry, the heat lost by the hot body = heat gained by a cold body

So, the heat lost by metals = heat gained by calorimeter

For calorimeter A, the heat lost by metal X = heat gained by the water

$\Rightarrow m{s}_X∆T=m{s}_w∆T$$\Rightarrow m{s}_X\left(T_1-T_A\right)=m{s}_w\left(T_A-T_W\right)$

$\Rightarrow 5g\times 0.2\frac{cal}{g^{o}C}\times \left({40}^{o}C-{22}^{o}C\right)=m_{w}g\times 1\frac{cal}{g^{o}C}\times \left({22}^{o}C-{20}^{o}C\right)$

$\Rightarrow 5\times 0.2\times 18=m_w\times 2$

$\Rightarrow m_w=5\times 0.2\times 9=9g$

Hence the mass of water is 9g.

Step 3: Finding specific heat of metal Y

For calorimeter B, the heat lost by metal Y = heat gained by the water

$\Rightarrow m{s}_Y∆T=m{s}_w∆T$$\Rightarrow m{s}_Y\left(T_1-T_B\right)=m{s}_w\left(T_B-T_W\right)$

$\Rightarrow 5g\times s_Y\times \left({40}^{o}C-{23}^{o}C\right)=9g\times 1\frac{cal}{g^{o}C}\times \left({23}^{o}C-{20}^{o}C\right)$

$\Rightarrow 5\times s_Y\times 17=9\times 3$

$\Rightarrow s_Y=\frac{3\times 9}{5\times 17}=\frac{27}{85}g$

Hence, the specific heat of metal Y is $\frac{27}{85}\frac{cal}{g^{o}C}$.