Question

A small electric immersion heater is used to heat $100\text{g}$ of water. The power on the heater is labelled as $200\text{W}$. Calculate the time required to raise the temperature from $23{}^{\circ }\text{C}$ to $100{}^{\circ }\text{C}$.
$[$Specific heat capacity of water is $4200\text{J/kg}{-}^{\circ }\text{C}]$

• A. $132\text{s}$
• B. $142\text{s}$
• C. $152\text{s}$
• D. $162\text{s}$

Answer 1

The correct option is D $162\text{s}$

Given:
Mass of water, $\left(m\right)=100\text{g}$ or $0.1\text{kg}$
Specific heat capacity of water, $\left(s\right)=4200\text{J/kg}{-}^{\circ }\text{C}$
Initial temperature, $\left(T_i\right)=23{}^{\circ }\text{C}$
Final temperature, $\left(T_f\right)=100{}^{\circ }\text{C}$
Power supplied by heater, $\left(P\right)=200\text{W}$
Let the time required be $t$.
Heat absorbed by water, $Q=ms(T_f-T_i)$
$\Rightarrow Q=0.100\times 4200\times (100-23)$
$\Rightarrow Q=32340\text{J}.......\left(1\right)$
Now, Power delivered $P=\frac{Q}{t}$
$\Rightarrow t=\frac{Q}{P}\Rightarrow t=\frac{32340}{200}=161.7\text{s}\approx 162\text{s}$
Hence, option $\left(d\right)$ is correct answer.