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})
-
A.
15minutes -
B.
20 minutes -
C.
42 minutes -
D.
50 minutes -
E.
84 minutes
Correct Answer: Option D
Explanation
H=pt =mL +mcθ ; 840t= 100×336+100×4.2×20 thus t=420000/840=50minutes