WAEC - Physics (1997 - No. 18)
A cup containing 100g of pure water at 20°C is placed in a refrigerator. If the refrigerator extracts heat at a rate of 840J per minute, calculate the time taken for the water to freeze. Neglect the heat capacity of the material of the cup. (Specific heat capacity of water =4.2J\(g^{-1}K^{-2}\))((Specific latent heat fusion of water =336J\(g^{-1}\)
15minutes
20 minutes
42 minutes
50 minutes
84 minutes
Explanation
H=pt =mL +mcθ ; 840t= 100x336+100x4.2x20 thus t=420000/840=50minutes
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