A small electric heater is used to heat $200 \mathrm{~g}$ of water. The time required to bring all this…
A small electric heater is used to heat $200 \mathrm{~g}$ of water. The time required to bring all this water from $40^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$ is $200 \mathrm{~s}$. If specific heat of the water is $4200 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$ then the power supplied by the heater is
$155 \mathrm{~W}$
$310 \mathrm{~W}$
$88 W$
$252 \mathrm{~W}$
Solution
Given, $t=200 \mathrm{~s}, m=200 \mathrm{~g}=0.2 \mathrm{~kg}$
$\begin{aligned} s & =4200 \mathrm{jkg}^{-1} \mathrm{~K}^{-1} \\ \Delta T & =(100+273)-(40+273)=60 \mathrm{~K}\end{aligned}$
Heat absorbed by water = Energy supplied by heater
$\Rightarrow$ Power of heater $\times$ time $=$ mass $\times$ specific heat $\times$ 'rise of temperature of water
$\begin{aligned} & \Rightarrow \quad P \times t=m \times s \times \Delta T \\ & \Rightarrow \quad P=\frac{m \times s \times \Delta T}{t} \\ & \Rightarrow \quad P=\frac{0.2 \times 4200 \times 60}{200}\end{aligned}$
$=252 \mathrm{~W}$