A 50 gm lead bullet specific heat 0.02 is initially at 30 degree Celsius.It is fired vertically upwards with a speed of 840 m/s and on returning to the starting level, strikes a cake of ice at 0 degree celsius.Hw much ice is melted?? Latent heat of ice = 80 cal/g.Assume there is no loss in energy. Explain??

Data:
Mass of lead bullet = m = 50 g = 50/1000 = 0.05 kg
Specific heat = 0.02 J
Initial temperature = 30°C
Speed = 840 m/s
Amount of melted ice = ?
Solution:
Total heat in a bullet = 50 x 0.02 x 30 
                                 = 30 cal
E = 1/2 mv² 
   = 1/2 (0.05) (840)²
   = 17640 J
Since, 1 J = 0.2389 cal
So, 17640 J = 4214.19 cal
Hence, E = 4214.19 cal 
Total heat = 4214.19 + 30 = 4244.19 J
Heat of fusion = 80 cal/g
So, total amount of melted ice = 4244.19 / 80 = 53 g
Hence, 53 g of ice is melted.