A calorimeter containing m = 1 kg of ice at a temperature of T1 = –10 ° C was added 200 g of water at a temperature

A calorimeter containing m = 1 kg of ice at a temperature of T1 = –10 ° C was added 200 g of water at a temperature of t2 = 100 ° C. Find the temperature t and the volume of the mixture V after the establishment of thermal equilibrium. Disregard the heat capacity of the calorimeter and heat losses.

ml = 1 kg.

tl = -10 ° C.

t1 = 0 ° C.

mw = 200 g = 0.2 kg.

tv = 100 ° C.

Cl = 2100 J / kg * ° С.

Cw = 4200 J / kg * ° С.

q = 3.4 * 10 ^ 5 J / kg.

ρ = 1000 kg / m3.

t -?

V -?

Heating of ice requires thermal energy: Q1 = Cl * ml * (t1 – tl).

Q1 = 2100 J / kg * ° C * 1 kg * (0 ° C + 10 ° C) = 21000 J.

When the water is cooled, thermal energy will be released: Q2 = Cw * mw * (tv – t1).

Q2 = 4200 J / kg * ° C * 0.2 kg * (1000 ° C – 0 ° C) = 84000 J.

To melt ice, the amount of heat is required: Q3 = q * ml.

Q3 = 3.4 * 10 ^ 5 J / kg * 1 kg = 340,000 J.

Only Q2 – Q1 = 84000 J – 21000 J = 63000 J.

ml “= 63000 J / 3.4 * 10 ^ 5 J = 0.19 kg.

Answer: the temperature will be set t1 = 0 ° С, there will be mw “= 0.39 kg of water and ml” “= 0.81 kg of ice.