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.
A
9.0kjkg\(^{-1}\)
B
84.0kjkg\(^{-1}\)
C
134.0kjkg\(^{-1}\)
D
168.0kjkg\(^{-1}\)
Verified Correct Answer: C.
Physical Principles & Numerical Solution
Heat involved (H) = 88,000J, Mass(M)= 0.5kg, specific heat capacity of water(C) = 4200JkgºC,
Specific latent heat of fusion(L) = ?, Temperature change(\Delta\theta) = \theta2 - \theta1 = (10 - 0)^\circ = 10^\circ
H = MC\Delta\theta + ML
or
H = M(C\Delta\theta + L) -->\(\frac{H}{M}\) = C\Delta\theta + L
: L = \(\frac{H}{M}\) - C\Delta\theta = \(\frac{88,000}{0.5}\) - 4200 \(\times\) 10
L = 176,000 - 42000 = 134,000
L = 134,000 or 134kj/kg
Source: WAEC / JAMB / NECO Classroom Archive (Record #60731)