0.5kg of water at 10ºC is completely converted to ice at 0ºC by extracting (88000) of heat from it. If the specific heat capacity of water is 4200jkg\(^{1}\) Cº. Calculate the specific latent heat of fusion of ice.
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Correct Answer: Option C
Explanation:
Heat involved (H) = 88,000J, Mass(M)= 0.5kg, specific heat capacity of water(C) = 4200JkgºC,
Specific latent heat of fusion(L) = ?, Temperature change(Δθ) = θ2 - θ1 = (10 - 0)° = 10°
H = MCΔθ + ML
or
H = M(CΔθ + L) -->\(\frac{H}{M}\) = CΔθ + L
: L = \(\frac{H}{M}\) - CΔθ = \(\frac{88,000}{0.5}\) - 4200 \(\times\) 10
L = 176,000 - 42000 = 134,000
L = 134,000 or 134kj/kg
Heat involved (H) = 88,000J, Mass(M)= 0.5kg, specific heat capacity of water(C) = 4200JkgºC,
Specific latent heat of fusion(L) = ?, Temperature change(Δθ) = θ2 - θ1 = (10 - 0)° = 10°
H = MCΔθ + ML
or
H = M(CΔθ + L) -->\(\frac{H}{M}\) = CΔθ + L
: L = \(\frac{H}{M}\) - CΔθ = \(\frac{88,000}{0.5}\) - 4200 \(\times\) 10
L = 176,000 - 42000 = 134,000
L = 134,000 or 134kj/kg