Test Index

JAMB Physics 2024 Paper

© africaexams.com

Question : 14

Total: 125

5 X 10

^{-3}

kg of liquid at its boiling point is evaporated in 20s by the heat generated by a resistor of 2

\Omega

when a current of 10A is used. The specific latent heat of vaporization of the liquid is

8.0 x 10 $^4$ Jkg $^{-1}$
8.0 x 10 $^5$ Jkg $^{-1}$
8.0 x 10 $^6$ Jkg $^{-1}$
8.0 x 10 $^7$ Jkg $^{-1}$

Solution:

Concept:

The electrical heat energy generated is entirely used to vaporize the liquid at its boiling point.

Explanation:

The heat produced by the resistor is given by

H = I^2 R t

.

Here,

I = 10\,A

,

R = 2\,\Omega

, and

t = 20\,s

.

So,

H = (10)^2 \times 2 \times 20 = 4000\,J

.

This heat equals the heat absorbed during vaporization:

H = m L

.

Given

m = 5 \times 10^{-3}\,kg

, we have

4000 = (5 \times 10^{-3})L

.

Therefore,

L = \frac{4000}{5 \times 10^{-3}} = 8.0 \times 10^5\,Jkg^{-1}

.

Answer:

Option B:

8.0 \times 10^5\,Jkg^{-1}

.

© africaexams.com

Go to Question:

Prev Question

Next Question