Question

An experiment takes $10\, min$ to raise the temperature of water in a container from $0^{\circ} C$ to $100^{\circ} C$ and another $55\, min$ to convert it totally into steam by a heater supplying heat at a uniform rate. Neglecting the specific heat of the container and taking specific heat of water to be $1\, cal / g { }^{\circ} C$, the heat of vaporisation according to this experiment will come out to be

• A. 530 cal/ g
• B. 540 cal/g
• C. 550 cal/g
• D. 560 cal/g

Answer 1

Answer: (C)

550 cal/g

Answer 2

Let $P$ be power of the heater
$P(10)= m \times 1 \times(100-0) ; P(55)=m\, L_{v}$
Dividing we get, $L_{v}=550\, cal / g$