In water weighing 5 kg at a temperature of 40 degrees, 1 kg of ice with a temperature of 0 degrees was lowered.

In water weighing 5 kg at a temperature of 40 degrees, 1 kg of ice with a temperature of 0 degrees was lowered. What will be the temperature of the resulting water after heat exchange?

mw = 5 kg.

tv = 40 ° C.

ml = 1 kg.

tl = 0 ° C.

Cw = 4200 J / kg * ° C.

λ = 3.4 * 105 J / kg.

t -?

To melt ice and heat the resulting water, the required amount of heat Q is expressed by the formula: Q = λ * ml + Cw * ml * (t – tl).

When the water cools down, the amount of heat will be released: Q = Cw * mwl * (tv – t).

λ * ml + Cw * ml * (t – tl) = Cw * mw * (tv – t).

λ * ml + Cw * ml * t – Cw * ml * tl = Cw * mw * tv – Cw * mw * t.

Since tl = 0 ° C, the heat balance equation will take the form: λ * ml + Cw * ml * t = Cw * mw * tv – Cw * mw * t.

Cw * ml * t + Cw * mw * t = Cw * mw * tv – λ * ml.

t = (Cw * mw * tv – λ * ml) / (Cw * ml + Cw * mw).

t = (4200 J / kg * ° C * 5 kg * 40 ° C – 3.4 * 105 J / kg * 1 kg) / (4200 J / kg * ° C * 1 kg + 4200 J / kg * ° C * 5 kg) = 19.8 ° C.

Answer: after heat exchange, the water temperature will be set at t = 19.8 ° C.