Question

find the time taken by a 500w heater to raise the temperature of 50kg of material of specific heat capacity 960Jper kg per kelvin from 18C to 38C.Assume that all the heat energy supplied by the heater is given to the material.

Answer 1

Given,
Power of heater=500 W=500 J/s
​Suppose the time for which the heater works is 't'.
Energy supplied by the heater, E=P x t=500 x t
$$\begin{gathered} \mathrm{Rise}\mathrm{in}\mathrm{temperature}\mathrm{of}\mathrm{the}\mathrm{material}\mathrm{is}\mathrm{given}\mathrm{by}, \\ ∆\mathrm{\theta }=38-18={20}^0\mathrm{C} \\ \mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{material},\mathrm{m}=50\mathrm{kg} \\ \mathrm{Specific}\mathrm{heat}\mathrm{capacity}\mathrm{of}\mathrm{the}\mathrm{material}\left(\mathrm{c}\right)=960{\mathrm{Jkg}}^{-1}{\mathrm{K}}^{-1} \\ \mathrm{Heat}\mathrm{absorbed}\mathrm{by}\mathrm{the}\mathrm{material}=\mathrm{mc}∆\mathrm{\theta } \\ =50\times 960\times 20=960000\mathrm{J} \\ \mathrm{Enegy}\mathrm{supplied}\mathrm{by}\mathrm{the}\mathrm{heater}=\mathrm{Heat}\mathrm{absorbed}\mathrm{by}\mathrm{the}\mathrm{material} \\ 500\times \mathrm{t}=960000 \\ \mathrm{or}\mathrm{t}=1920\mathrm{s}\mathrm{or}32\min \end{gathered}$$