JAMB - Physics (2017 - No. 8)
Calculate the specific latent heat of vaporization of steam of 1.13 x 106J of heat energy is required to convert 15kg of steam to water.
7.53 x 105 Jkg\(^{-1}\)
7.53 x 10-2Jkg\(^{-1}\)
\(7.53 \times 10^{4}\)Jkg\(^{-1}\)
7.53 x 10-3Jkg\(^{-1}\)
Explanation
Quantity of heat = mass(m) × specific latent heat (L)
Q = mL
L = \(\frac{Q}{m}\)
= \(\frac{1.13 \times 10^{6}}{15}\)
= \(7.53 \times 10^{4}\)Jkg\(^{-1}\)
Q = mL
L = \(\frac{Q}{m}\)
= \(\frac{1.13 \times 10^{6}}{15}\)
= \(7.53 \times 10^{4}\)Jkg\(^{-1}\)
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