Question

An experiment takes $10\text{min}$ for an electric kettle to heat a certain quantity of water from $0^{\circ }\text{C}$ to ${100}^{\circ }\text{C}$. If it takes $54\text{min}$ to convert this water at ${100}^{\circ }\text{C}$ into steam, then latent heat of steam is:

• A. $80\text{cal/gm}$
• B. $540\text{cal/kg}$
• C. $540\text{cal/gm}$
• D. $80\text{cal/kg}$

Answer 1

The correct option is C $540\text{cal/gm}$

Let $m$ be the mass of water.
Quantity of heat absorbed by water in $10\text{min}$
$=mc\Delta T=m\times 1\times 100=100m$ (in $\text{calories}$)
$\therefore$ Quantity of heat absorbed by water in $54\text{min}=\frac{100m\times 54}{10}$
Quantity of heat required to convert water into steam $=mL$
Hence, $\frac{100m\times 54}{10}=\text{mL}$
or $L=540\text{cal/gm}$