Question

An electric immersion heater is switched on for$8minutes$. The heat supplied by it raises the temperature of $500g$ of water from $10°C$to $60°C$. Calculate the power of the heater in watts. [Specific heat of water$=\text{4.2}{\mathrm{Jg}}^{-1}{}^{\circ }\mathrm{C}^{-1}$]

Answer 1

Step 1: Given data

Time $\left(t\right)=8minutes=480seconds$

Mass$\left(m\right)=500g$

Temperature changes from $10°C$ to $60°C$

Specific heat of water$\left(c\right)=\text{4.2}{\mathrm{Jg}}^{-1}{}^{\circ }\mathrm{C}^{-1}$

Power (P)=?

Step 2: Calculating the heat energy absorbed

Rise in temperature $\left({\theta }_R\right)=(60-10)°C=50°C$

Heat energy absorbed$=mc{\theta }_R$

$$\begin{gathered} =500g\times \text{4.2}{\mathrm{Jg}}^{-1}{}^{\circ }\mathrm{C}^{-1}\times 50°\mathrm{C} \\ =105000\mathrm{J} \end{gathered}$$

Heat energy= product of power and time

$$\begin{gathered} \Rightarrow P\times 480s=105000J \\ \Rightarrow P=\frac{105000J}{480s} \\ \Rightarrow P=218.75watt \end{gathered}$$

The power of the heater is $218.75watt$.