A piece of ice is placed in a calorimeter containing 500 g of water at a temperature of 20 ° C at a temperature of 0 ° C.

A piece of ice is placed in a calorimeter containing 500 g of water at a temperature of 20 ° C at a temperature of 0 ° C. What is the smallest mass of ice needed to bring the temperature of the calorimeter contents to 0 ° C?

Given:
mw = 500 g = 0.5 kg
tv = 20 ° С
tl = 0 ° С
tcon = 0 ° С
sv = 4200 J / kg • ° C
λl = 3.3 • 105 J / kg
Find: ml -?
Decision
Qw + Ql = 0. Qw = cmw (t2w – t1w) – the amount of heat given off by the water. Ice is taken at the melting point, which means:
Ql = λlml is the amount of heat received by ice. Therefore:
cmv (t2v – t1v) + λlml = cmv (t2v – t1v), or λlml = cmv (t2v – t1v)
ml = (svmv (tv – tcon)) / λl = 4200 * 0.5 (20 – 0) / 3.3 * 105 = 42000/330000 = 0.127 g