Question

An electric immersion heater of $1.08\text{kW}$ is immersed in water. After it has reached a temperature of ${100}^{\circ }C$ , how much time will be require to produce $100\text{g}$ of steam ?

• A. $50\text{s}$
• B. $420\text{s}$
• C. $105\text{s}$
• D. $210\text{s}$

Answer 1

The correct option is D $210\text{s}$

We know that,
$L=\frac{Q}{m}$
$Q=mL$
$Pt=mL$
Here, $Q=$ amount of heat absorbed or released
$m=$ mass
$P=$ power consumed
$t=$time
$L=$Specific latent heat coefficiant of the material $\frac{J}{kg}$

Time required to boil $100\text{g}$ water
$t=540\times 4.2\left(\frac{100}{1080}\right)=210\text{s}$