In a vessel with water weighing 2 kg, having a temperature of 30 ° C, ice was placed with a mass of 1 kg

In a vessel with water weighing 2 kg, having a temperature of 30 ° C, ice was placed with a mass of 1 kg, the temperature of which is 0 ° C. What is the temperature in the vessel?

mw = 2 kg.

tv = 30 ° C.

ml = 1 kg.

tl = 0 ° C.

Cw = 4200 J / kg * ° C.

λ = 340,000 J / kg.

t -?

The required amount of heat for melting ice, Ql, is expressed by the formula: Ql = λ * ml, where λ is the specific heat of melting of ice, ml is the mass of ice.

Ql = 340,000 J / kg * 1 kg = 340,000 J.

The amount of heat Qw, which will be released when the water cools down, can be expressed by the formula: Qw = Cw * mw * (tw – t), where Cw is the specific heat capacity of water, mw is the mass of water, t, tw are the final and initial water temperatures.

Qw = 4200 J / kg * ° C * 2 kg * (30 ° C – 0 ° C) = 252000 J.

Since the amount of heat that will be released when the water cools is not enough to melt all the ice, we will find how much ice can be melted when the water cools down.

Ql “= λ * ml”, where ml “is the mass of ice that will melt.

Ql “= Qv.

ml “= Qw / λ.

ml “= 252000 J / 340000 J / kg = 0.74 kg.

Answer: the temperature in the vessel will be t = 0 ° C and it will contain ml = 0.26 kg of ice and mw = 2.74 kg of water.