Question on: JAMB Physics - 2017
Calculate the specific latent heat of vaporization of steam of 1.13 x 106J pf heat energy is required to convert 15kg of it to water.
A
7.53 x 105 Jkg\(^{-1}\)
B
7.53 x 10-2Jkg\(^{-1}\)
C
7.53 x 104Jkg\(^{-1}\)
D
7.53 x 10-3Jkg\(^{-1}\)
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Correct Option: C
Latent Energy = mass(m) × specific latent heat (L)
ε = mL
L = ε/m\(\frac{E}{m}\)
= \(\frac{1.13 \times 106}{15}\)
= 7.53 × 104Jkg\(^{-1}\)
ε = mL
L = ε/m\(\frac{E}{m}\)
= \(\frac{1.13 \times 106}{15}\)
= 7.53 × 104Jkg\(^{-1}\)
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