Question

An electric immersion heater of $1.08kW$ is immersed in water. After the water has reached a temperature of ${100}^{o}C$, how much time will be required to produce $100g$ of steam

• A. $50s$
• B. $420s$
• C. $105s$
• D. $210s$

Answer 1

Answer: (D)

$210s$

L is the latent heat of vaporization of water, the heat required for producing 1 g of steam.

So, $L=540cal=540\times 4.2=2268J$

Here, $mL=Pt$ where $P=1.08kW$ is the supplied power and the required time.

thus, $t=\frac{mL}{P}=\frac{100\times 2268}{1.08\times {10}^3}=210s$