There is not enough ice to survive, so all of it will melt. The specfic heat of liquid ice is 4186 H/kg C.
The heat absorbed by the ice in melting and rising to temperature T must equal the heat lost by the water in cooling from 50 to the final T.
1kg*(336,000 + 4186T) = 9kg*4186*(50-T)
336,000 = 4186*[9(50-T)-T]
80.3 = 450 - 10T
10T = 369.7
T = 37.0 C (rounded off)
if 1 kilogram of ice is dropped in 9 kilogram of water at 50 degree centigrade. what will be final temperature of water. while the latent specific heat of fusion of ice is 336000 joule/kilogram
1 answer
1 answer