150 g of water was poured into the beaker of the calorimeter. The initial temperature of the calorimeter and water is 55C

150 g of water was poured into the beaker of the calorimeter. The initial temperature of the calorimeter and water is 55C. A piece of ice with a temperature of 0C was lowered into the water. After thermal equilibrium was reached, the water temperature in the calorimeter became 5C. Determine the ice mass; ignore the heat capacity of the calorimeter.

mw = 150 g = 0.15 kg.

Cw = 4200 J / kg * ° C.

λ = 3.4 * 10 ^ 5 J / kg.

t1 = 55 ° C.

t2 = 0 ° C.

t = 5 ° C.

ml -?

When the water was cooled, the amount of heat Q was released, which we express by the formula: Q = Cw * mw * (t1-t), where Cw is the specific heat capacity of water, mw is the mass of water that has cooled down, t1, t is the initial final temperature of the water.

Since ice is at the melting temperature, the amount of heat Q, which is necessary for its melting and heating of the resulting water, is expressed by the formula: Q = λ * ml + Cw * ml * (t – t2), where Cw is the specific heat capacity of water, ml – ice mass, t1, t – initial and final water temperature.

Cw * mw * (t1- t) = λ * ml + Cw * ml * (t – t2).

ml = Cw * mw * (t1- t) / (λ + Cw * (t – t2)).

ml = 4200 J / kg * ° C * 0.15 kg * (55 ° C – 5 ° C) / (3.4 * 10 ^ 5 J / kg + 4200 J / kg * ° C * (5 ° C – 0 ° C)) = 0.087 kg.

Answer: the mass of ice is ml = 0.087 kg.