A vessel of negligible heat capacity contains 0.20\text{ kg} of a liquid at its boiling po… — JAMB UTME

A vessel of negligible heat capacity contains $0.20\text{ kg}$ of a liquid at its boiling point of $100^\circ\text{C}$ . An electric heating element rated at $500\text{ W}$ is immersed in the liquid and switched on for $3.0\text{ minutes}$ . If $0.040\text{ kg}$ of the liquid is vaporized during this period, what is the specific latent heat of vaporization of the liquid?

$2.25 \times 10^6\text{ J kg}^{-1}$ Answer
B
$2.25 \times 10^4\text{ J kg}^{-1}$
C
$4.50 \times 10^5\text{ J kg}^{-1}$
D
$1.25 \times 10^4\text{ J kg}^{-1}$

Answer

The specific latent heat of vaporization of the liquid is

2.25 \times 10^6\text{ J kg}^{-1}

The electrical energy supplied by the heater in

180\text{ seconds}

Q = 500 \times 180 = 90,000\text{ J}

. Since phase change occurs at constant boiling temperature (

100^\circ\text{C}

), the heat required to vaporize

0.040\text{ kg}

of liquid is given by

Q = m L_v

. Solving for

L_v

yields

\frac{90,000\text{ J}}{0.040\text{ kg}} = 2.25 \times 10^6\text{ J kg}^{-1}

Step-by-Step Solution

Convert the time duration into seconds

t = 3.0\text{ minutes} \times 60\text{ s/min} = 180\text{ s}

SI units require time to be in seconds when calculating electrical energy.

Calculate total electrical thermal energy supplied by the heater

Q = P \times t = 500\text{ W} \times 180\text{ s} = 90,000\text{ J}

Thermal energy produced equals power multiplied by time elapsed.

Apply the latent heat formula using the mass of liquid vaporized

L_v = \frac{Q}{m} = \frac{90,000\text{ J}}{0.040\text{ kg}} = 2,250,000\text{ J kg}^{-1} = 2.25 \times 10^6\text{ J kg}^{-1}

During vaporization at constant boiling temperature, heat energy supplied goes entirely into changing state.

Key Concept

Specific Latent Heat of Vaporization

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