The specific latent heat of fusion of an object is 2.5 times smaller than its specific heat capacity. 250g of the solid object is heated from its freezing point to 20 K above its freezing point by a 125 W heater applied for 8 minutes. Calculate both the specific latent heat of fusion and the specific heat capacity of the object.
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Answer:
Let the specific latent heat of fusion = l
thus, the specific heat capacity = 2.5 l
Total energy = Pt = ml + mcΔθ
125 * 8 * 60 = 0.25 * l + 0.25 * 2.5 l * 20
60000 = 0.25 l + 12.5 l
12.75 l = 60000
Thus, specific latent heat of fusion, l = 4705.9 J/kg
Specific heat capacity = 2.5 l = 11764.7 J /kg K
