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 =4.2Jg−1K−2)((Specific latent heat fusion of water =336Jg−1
A. 15minutes
B. 20 minutes
C. 42 minutes
D. 50 minutes
E. 84 minutes
Correct Answer:
Option D – 50 minutes
Explanation
H=pt =mL +mcθ ; 840t= 100×336+100×4.2×20 thus t=420000/840=50minutes
