Question

200 g mass of a certain metal at 83°C is immersed in 300 g of water at 30°C. The final temperature becomes 33°C. Calculate the specific heat capacity of the metal.

Answer 1

Heat lost by metal = Heat gained by water
$200 \times c \times (83-33) = 300 \times 4.2 \times (33-30)$

Hence c = 0.378 J/(g °C)

Answer 2

Answer:

200 g mass of a certain metal at 83°C is immersed in 300 g of water at 30°C. The final temperature becomes 33°C. The specific heat capacity of the metal will be $0.372 \ \mathrm{Jg}^{-1} \mathrm{K}^{-1}$

Explanation:

• Let M be the mass of water, m be the mass of the solid, T be the temperature of the water, t be the temperature of the solid, and θ be the final temperature.

• We can solve this problem using the principle of calorie meter.

       According to the principle of calorie meter,

       Heat gained by water =Heat lost by the metal

       That is,

       $\mathrm{MC}_{1}(\mathrm{~T}-\theta)=m \mathrm{C}_{2}(\theta-t)$

       Where

       $C_1$ - Specific heat capacity of water

       $C_2$ - Specific heat capacity of metal

In the question, it is given that,

$M = 200 g$          $m=300 \mathrm{~g}$    

$T = 83^0C$          $t = 30^0C$    

$\theta = 33^0C$

We know,        

$C_1 = \text { 4.2 Joule } \mathrm{g}^{-1} \mathrm{~K}^{-1}$

Substitute these values into the above equation,

$300 \times 4.2(30-33)=200 \times \mathrm{C}_{\mathrm{2}} \times(33-83)$

$C_{2}=\frac{(-3720) }{(-200 \times 50)} =0.372 \ \mathrm{Jg}^{-1} \mathrm{K}^{-1}$