Question

A solid of mass 50 g at 150°C is placed in 100 g of water at 110°C, when the final temperature recorded is 200°C. Find the specific heat capacity of the solid (Specific heat capacity of water 4.2 J g -1°C -1).

Answer 1

Explanation:

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y - 1 = z

sin(x-y)= sinxcosy- cosxsiny

Answer 2

Answer:

A solid of mass 50 g at 150°C is placed in 100 g of water at 110°C when the final temperature recorded is 200°C. The specific heat capacity of the solid will be $-15.12 \mathrm{Jg}^{-1} \mathrm{~K}^{-1}$

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,

        $\begin{array}{ll}M=100 \mathrm{~g} & m=50 \mathrm{~g} \\T=110^{0} \mathrm{C} & t=150^{\circ} \mathrm{C} \\\theta=200^{0} \mathrm{C} &\end{array}$

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

        Substitute these values into the above equation,

        $\begin{array}{l}100 \times 4.2(110-200)=50 \times \mathrm{C}_{2} \times(200-150) \\C_{2}=\frac{(-37800)}{(50 \times 50)}=-15.12 \mathrm{Jg}^{-1} \mathrm{~K}^{-1}\end{array}$