Question

A brass rod of 0.2 kg at 100⁰C is dropped into 0.5 kg of water at 20⁰C. The final temperature of the mixture is 23⁰C. Calculate the specific heat capacity of brass(given specific heat capacity of water= 1cal/gm⁰C).

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{The}\mathrm{water}\mathrm{will}\mathrm{heated}\mathrm{while}\mathrm{the}\mathrm{rod}\mathrm{brass}\mathrm{will}\mathrm{be}\mathrm{cooled}.\mathrm{We}\mathrm{can}\mathrm{write}\mathrm{the}\mathrm{heat}\mathrm{balance}\mathrm{equation}, \\ {\mathrm{m}}_{\mathrm{w}}{\mathrm{c}}_{\mathrm{w}}\left(\mathrm{T}-{\mathrm{t}}_{\mathrm{w}}\right)={\mathrm{m}}_{\mathrm{b}}{\mathrm{c}}_{\mathrm{b}}\left({\mathrm{t}}_{\mathrm{b}}-\mathrm{T}\right) \\ \mathrm{where},{\mathrm{m}}_{\mathrm{w}}=\mathrm{mass}\mathrm{of}\mathrm{water} \\ {\mathrm{c}}_{\mathrm{w}}=\mathrm{specific}\mathrm{heat}\mathrm{capacity}\mathrm{of}\mathrm{water} \\ {\mathrm{t}}_{\mathrm{w}}=\mathrm{temperature}\mathrm{of}\mathrm{water} \\ {\mathrm{m}}_{\mathrm{b}}=\mathrm{mass}\mathrm{of}\mathrm{brass} \\ {\mathrm{c}}_{\mathrm{b}}=\mathrm{specific}\mathrm{heat}\mathrm{capacity}\mathrm{of}\mathrm{brass} \\ {\mathrm{t}}_{\mathrm{b}}=\mathrm{temperature}\mathrm{of}\mathrm{brass} \\ \mathrm{T}=\mathrm{final}\mathrm{temperature}\mathrm{of}\mathrm{mixture} \\ {\mathrm{c}}_{\mathrm{w}}=4200\mathrm{J}/\mathrm{kg}°\mathrm{C} \\ {\mathrm{c}}_{\mathrm{b}}=\frac{{\mathrm{m}}_{\mathrm{w}}{\mathrm{c}}_{\mathrm{w}}\left(\mathrm{T}-{\mathrm{t}}_{\mathrm{w}}\right)}{{\mathrm{m}}_{\mathrm{b}}\left({\mathrm{t}}_{\mathrm{b}}-\mathrm{T}\right)} \\ {\mathrm{c}}_{\mathrm{b}}=\frac{0.5\times 4200\left(23°-20°\right)}{0.2\times \left(100°-23°\right)} \\ {\mathrm{c}}_{\mathrm{b}}=409.09\mathrm{J}/\mathrm{kg}°\mathrm{C} \end{gathered}$$