The calorimeter contains ice and water at 0 * C. The mass of ice and water is the same and equal to 0.5 kg.

The calorimeter contains ice and water at 0 * C. The mass of ice and water is the same and equal to 0.5 kg. Water weighing 1 kg is poured into the calorimeter at a temperature of 50 * C. What temperature will be established in the calorimeter after heat exchange?

ml = 0.5 kg.

mw1 = 0.5 kg.

mv2 = 1 kg.

t1 = 0 ° C.

t2 = 50 ° C.

Cw = 4200 J / kg * ° С.

q = 3.4 * 10 ^ 5 J / kg.

t -?

Let us express the amount of heat Q that warm water will give during cooling: Q = Sv * mw2 * (t2 – t).

Let us express the amount of heat Q, which goes to melting ice and heating water: Q = q * ml + Sv * ml * (t – t1) + Sv * mw1 * (t – t1).

Let us write down the heat balance equation: Sv * mw2 * (t2 – t) = q * ml + Sv * ml * (t – t1) + Sv * mw1 * (t – t1).

Sv * mv2 * t2 – Sv * mv2 * t = q * ml + Sv * ml * t – Sv * ml * t1 + Sv * mv1 * t – Sv * mv1 * t1.

Sv * mv2 * t2 – q * ml + Sv * ml * t1 + Sv * mv1 * t1 = Sv * ml * t + Sv * mv1 * t + Sv * mv2 * t.

t = (Sv * mv2 * t2 – q * ml + Sv * ml * t1 + Sv * mv1 * t1) / (Sv * ml + Sv * mv1 + Sv * mv2).

t = (Sv * mv2 * t2 – q * ml + Sv * ml * t1 + Sv * mv1 * t1) / (Sv * ml + Sv * mv1 + Sv * mv2).

t = (4200 J / kg * ° C * 1 kg * 50 ° C – 3.4 * 10 ^ 5 J / kg * 0.5 kg + 4200 J / kg * ° C * 0.5 kg * 0 ° C + 4200 J / kg * ° C * 0.5 kg * 0 ° C) / (4200 J / kg * ° C * 0.5 kg + 4200 J / kg * ° C * 0.5 kg + 4200 J / kg * ° С * 1 kg) = 4.8 ° С.

Answer: the temperature will be set in the calorimeter t = 4.8 ° С.