JAMB - Physics (2024 - No. 14)
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\(^{-1}\)
Explanation
Electrical heat generated = quantity of heat required to change water from liquid to vapour
I\(^2\) R t = mL
I = current = 10A, R = resistance = 2Ω, t = time = 20s, m = mass of water = 5 X 10\(^{-3}\)kg and L = specific latent heat of vaporization.
10\(^2\) x 2 x 20 = 5 X 10\(^{-3}\)L
4000 = 5 X 10\(^{-3}\)L
L = \(\frac{4000}{5\times10^{-3}}\) = 800000 ⇔ 8.0 x 10\(^5\)Jkg\(^{-1}\)
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