Question

an electric heater raises the temperature of 5000 g of agiven liquid . the electric heater is rated 1kW power. To raise the temperature of the liquid from 25degrees to 31degree , a heater requires 120 seconds. Calculate (i) the heat capacity of the liquid
(ii) the specific heat capacity of the liquid

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{temperature},\mathrm{t}=25°\mathrm{C}\mathrm{to}31°\mathrm{C} \\ \mathrm{mass},\mathrm{m}=5000\mathrm{g}=5\mathrm{kg} \\ \mathrm{time},\mathrm{T}=120\mathrm{s} \\ \mathrm{Power},\mathrm{P}=1\mathrm{kW} \\ \mathrm{Heat}\mathrm{capacity}=\frac{\mathrm{amount}\mathrm{of}\mathrm{heat}}{\mathrm{rise}\mathrm{in}\mathrm{temperature}} \\ \mathrm{electrical}\mathrm{energy}\mathrm{supplied},\mathrm{E}=\mathrm{P}\times \mathrm{T} \\ \mathrm{E}=1000\mathrm{W}\times 120\mathrm{s}=120000\mathrm{J} \\ \mathrm{It}\mathrm{is}\mathrm{in}\mathrm{joule}\mathrm{because}\mathrm{electrical}\mathrm{energy}\mathrm{is}\mathrm{supplied}\mathrm{is}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\mathrm{heat}. \\ \mathrm{Heat}\mathrm{capacity}=\frac{120000\mathrm{J}}{31°\mathrm{C}-25°\mathrm{C}} \\ \mathrm{Heat}\mathrm{capacity}=\frac{120000\mathrm{J}}{6°\mathrm{C}} \\ \mathrm{Heat}\mathrm{capacity}=20\times {10}^3\mathrm{J}/°\mathrm{C} \\ \mathrm{Specific}\mathrm{heat}\mathrm{capacity}=\frac{\mathrm{Heat}\mathrm{capacity}}{\mathrm{mass}} \\ \mathrm{Specific}\mathrm{heat}\mathrm{capacity}=\frac{20\times {10}^3}{5000} \\ \mathrm{Specific}\mathrm{heat}\mathrm{capacity}=4{\mathrm{Jkg}}^{-1}{\mathrm{C}}^{-1} \end{gathered}$$