Question

A solid of mass 50 g at 150°C is placed in 100 g of water at 10°C, when the final temperature recorded is 20°C. Find the specific heat capacity of the solid. (Specific heat capacity of water = 4.2 J g-1 °C-1)

please solve it​

Answer 1

Explanation:

Let subscript 1 denote solid (hot body) and subscript 2 denote water (cold body)

Initial temp. of solid T₁-150°C

Mass of solid m₁ = 50 g = 0.05 Kg Initial temp. of water T₂=11°C

Mass of water m₂= 100 g = 0.1 Kg

Final temp of both T= 20°C Specific heat capacity of water c₂=4.2 J/

g°C

Let Specific heat capacity of solid be c₁

now,

heat gained by water = heat lost by solid heat gained/lost = mcAT

.. m₁c₁AT₁ = m₂C₂AT2

→ 0.05*c₁*(150-20) = 0.1*4.2*(20-11)

0.05*130*c₁= = 0.1*4.2*9

⇒ C₁ = 0.5815 J/g°C

Answer: Specific heat capacity of the solid is 0.5815 J/g°C

Answer 2

Answer:

A solid of mass 50 g at 150°C is placed in 100 g of water at 10°C when the final temperature recorded is 20°C. The specific heat capacity of the solid will be $0.6461 \ \mathrm{Jg}^{-1} \mathrm{K}^{-1}$

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 solid

       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 solid

In the question, it is given that,

$M = 100 g$          $m=50 \mathrm{~g}$    

$T = 10^0C$           $t = 150^0C$    

$\theta = 20^0C$            $C_1 = \text { 4.2 Joule } \mathrm{g}^{-1} \mathrm{~K}^{-1}$

Substitute these values into the above equation,

$100 \times 4.2(10-20)=50 \times \mathrm{C}_{\mathrm{2}} \times(20-150)$

$C_{2}=\frac{(-4200) }{(-130 \times 50)} =0.6461 \ \mathrm{Jg}^{-1} \mathrm{K}^{-1}$