0.5kg of water at 10ºC is completely converted to ic...
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.
9.0kjkg\(^{-1}\)
84.0kjkg\(^{-1}\)
134.0kjkg\(^{-1}\)
168.0kjkg\(^{-1}\)
Correct answer is C
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