The jar contains 0.5 liters of hot water at a temperature of 70 ° C. What will the water temperature
The jar contains 0.5 liters of hot water at a temperature of 70 ° C. What will the water temperature be if you throw 100 g of ice taken at a temperature of -20 ° C into it? Heat exchange with ambient air is neglected.
Let us estimate the amount of heat required for cooling and crystallization of a liquid. From the condition of the problem it is known that the bank contains V = 0.5 l = 0.0005 m³ of hot water at a temperature of t₁ = 70 ° C. From the reference books we find: density of water ρ = 1000 kg / m³; specific heat capacity of water c₁ = 4200 J / (kg ° C); crystallization (melting) temperature of water tо = 0 ° С; specific heat of crystallization (melting) of ice λ = 340,000 J / kg. We get:
m₁ = ρ ∙ V; m₁ = 1000 kg / m³ ∙ 0.0005 m³; m₁ = 0.5 kg is the mass of hot water in the bank;
Q₁ = m₁ ∙ с₁ ∙ (t₁ – to); Q₁ = 0.5 kg ∙ 4200 J / (kg ° С) ∙ (70 ° С – 0 ° С); Q₁ = 147000 J – the amount of heat given off during cooling to 0 ° C.
Q₂ = λ ∙ m; Q₂ = 0.5 kg ∙ 340,000 J / kg; Q₂ = 170,000 J – the amount of heat given off during crystallization.
Absorbed amount of heat
Let us estimate the amount of heat required for heating and melting m₂ = 100 g = 0.1 kg of ice taken at a temperature of t₂ = – 20 ° C. From the reference books we find: the specific heat capacity of ice c₂ = 2100 J / (kg ° C). We get:
Q₃ = m₂ ∙ с₂ ∙ (to – t₂); Q₃ = 0.1 kg ∙ 2100 J / (kg ° С) ∙ (0 ° С – (- 20 ° С)); Q₃ = 4200 J is the amount of heat absorbed when ice is heated to 0 ° C.
Q₄ = m₂ ∙ λ; Q₄ = 0.1 kg ∙ 340,000 J / kg; Q₄ = 34000 J is the amount of heat absorbed during melting.
Heat balance equation
Comparing Q₁, Q₂, Q₃ and Q₄, we come to the conclusion that ice, heating up and taking heat from water, will increase its temperature to the melting point of the substance and turn into a liquid entirely. Further, these liquids will equalize the temperature to the value of t due to heat exchange. We compose the heat balance equation, in which the law of conservation of energy should be taken into account, that is, the heat received by a cold body is equal to the heat given off by a hot body:
m₁ ∙ с₁ ∙ (t₁ – t) = Q₃ + Q₄ + m₂ ∙ с₁ ∙ (t – to);
0.5 kg ∙ 4200 J / (kg ° C) ∙ (70 ° C – t) = 4200 J + 34000 J + 0.1 kg ∙ 4200 J / (kg ° C) ∙ (t – 0 ° FROM);
t = 43.1746 ° C ≈ 43.2 ° C.
Answer: the water temperature will become ≈ 43.2 ° С.
