Question

A  metal of mass 300gm is heated to a temperature 700 C.. It is then placed in 100gm of water at 250C . The  final steady temperature of water becomes     30o C . Neglecting the heat taken by the container , calculate the specific heat capacity of metal. ( specific heat capacity of water =4.2 J/g K )​

Answer 1

Answer:

Let's assume a constant value (c) for the Specific Heat of water, between freezing and boiling temperatures and at atmospheric pressure. After mixing the water, the ...

Answer 2

Answer:

A  metal of mass 300gm is heated to a temperature $70^0C$ It is then placed in 100gm of water at $25^0C$ The final steady temperature of the water becomes     $30^0C$ then the  specific heat capacity of metal will be $0.175 \mathrm{Jg}^{-1} \mathrm{~K}^{-1}$

Explanation:

• 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

• Consider a metal having mass m, temperature t, and specific heat capacity $C_2$ Let M be the mass of water, T be the temperature of the water, and $C_1$ be the specific heat capacity of water.

        θ, be the final temperature.

        That is,

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

     

• In the question, it is given that,

        $\begin{array}{ll}\ M=100 \mathrm{~g} & m=300 \mathrm{~g} \\\ T=25^{0} \mathrm{C} & t=70^{\circ} \mathrm{C} \\\theta=30^{0} \mathrm{C} &\end{array}$

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

        Substitute these values into the above equation,

        $\begin{array}{l}100 \times 4.2(25-30)=300 \times \mathrm{C}_{2} \times(30-70) \\C_{2}=\frac{-2100)}{(300 \times- 40)}=0.175 \mathrm{Jg}^{-1} \mathrm{~K}^{-1}\end{array}$