Question

В термосе находится кипяток, заполняющий термос наполовину. В медленно термос засыпают мокрый снег, состоящий на 55 % (по объему) из кристаллов льда и на 45% из воды, находящихся в равновесии, до тех пор, пока термос не будет заполнен полностью. Какая температура будет у содержимого термоса после установления равновесия? Используйте следующие значения характеристик веществ: плотность льда равна 900 кг/м3, плотность воды – 1000 кг/м3, удельная теплота плавления льда λ = 336 кДж/кг, удельная теплоемкость воды c = 4,2 кДж/(кг·°С). Теплообменом с окружающими телами и теплоемкостью колбы термоса пренебречь. Ответ запишите в градусах по шкале Цельсия, с точностью до десятых, без указания единиц измерения.

Answer 1

Ответ:

Vт - объем термоса

рл - плотность льда

рв - плотность воды

j - уд. теплота плавления льда

tр - равновесная температура

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Масса снегового льда

mл=Vл*pл=0,6(0,5Vт) *рл=0,3Vт*рл

Масса снеговой воды

mc=Vв*рв=0,4(0,5Vт) *рв=0,2Vт*рв

На таяние льда пошла теплота

Q2=j*mл=j*0,3Vт*рл

Образовалась"нулевая" вода. Ее масса

mo=mс + mл=Vт (0,2*рв + 0,3*рл)

Объем "нулевой" воды

Vo=mo/pв=Vт*(0,2рв + 0,3рл) /рв

Объем кипятка

V100=Vт - Vo=Vт*[1 - (0,2рв + 0,3рл) /рв] =Vт (0,8рв-0,3рл) /рв

Масса кипятка

m100=рв*V100=Vт (0,8рв - 0,3рл)

Теплота, отданная кипятком

Q1=c*m100*(100 - tp)

Теплота, переданная "нулевой" воде

Q3=c*mo*(tp - 0)=c*mo*tp

Q1 = Q2+Q3

Отсюда находим tp. Преобразования не привожу

tр=[c*(80рв + 30рл) - 0,3*j*рл] /(c*рв) = (222,6 - 90,72)/4,2= 31,4 град.

Объяснение:

Answer 2

Problem Analysis

We are given a thermos that is initially filled halfway with boiling water. Slowly, wet snow is added to the thermos, which consists of 55% ice crystals and 45% water by volume. We need to determine the temperature of the contents of the thermos after equilibrium is reached. We are provided with the following substance characteristics: the density of ice is 900 kg/m³, the density of water is 1000 kg/m³, the specific heat of fusion of ice (latent heat) is 336 kJ/kg, and the specific heat capacity of water is 4.2 kJ/(kg·°C).

Solution

To solve this problem, we need to consider the heat transfer that occurs during the process. The heat transfer can be divided into two parts: the heat transfer required to melt the ice and the heat transfer required to raise the temperature of the water.

Let's assume that the final temperature of the contents of the thermos is T°C.

Heat Transfer to Melt the Ice

The heat transfer required to melt the ice can be calculated using the formula:

Q₁ = m₁ * λ₁

Where: - Q₁ is the heat transfer required to melt the ice (in joules) - m₁ is the mass of the ice (in kg) - λ₁ is the specific heat of fusion of ice (in kJ/kg)

The mass of the ice can be calculated using the density of ice and the volume of the ice:

m₁ = ρ₁ * V₁

Where: - ρ₁ is the density of ice (in kg/m³) - V₁ is the volume of the ice (in m³)

The volume of the ice can be calculated using the volume of the snow and the percentage of ice crystals in the snow:

V₁ = V_snow * (55/100)

Where: - V_snow is the volume of the snow (in m³)

Heat Transfer to Raise the Temperature of the Water

The heat transfer required to raise the temperature of the water can be calculated using the formula:

Q₂ = m₂ * c₂ * ΔT

Where: - Q₂ is the heat transfer required to raise the temperature of the water (in joules) - m₂ is the mass of the water (in kg) - c₂ is the specific heat capacity of water (in kJ/(kg·°C)) - ΔT is the change in temperature of the water (in °C)

The mass of the water can be calculated using the density of water and the volume of the water:

m₂ = ρ₂ * V₂

Where: - ρ₂ is the density of water (in kg/m³) - V₂ is the volume of the water (in m³)

The volume of the water can be calculated using the volume of the snow and the percentage of water in the snow:

V₂ = V_snow * (45/100)

Where: - V_snow is the volume of the snow (in m³)

Equating the Heat Transfers

Since the heat transfer required to melt the ice and the heat transfer required to raise the temperature of the water are both supplied by the boiling water, they must be equal. Therefore, we can equate the two heat transfers:

Q₁ = Q₂

Substituting the expressions for Q₁ and Q₂, we get:

m₁ * λ₁ = m₂ * c₂ * ΔT

Substituting the expressions for m₁, m₂, V₁, and V₂, we get:

(ρ₁ * V₁) * λ₁ = (ρ₂ * V₂) * c₂ * ΔT

Substituting the expressions for V₁ and V₂, we get:

(ρ₁ * V_snow * (55/100)) * λ₁ = (ρ₂ * V_snow * (45/100)) * c₂ * ΔT

Simplifying the equation, we get:

(ρ₁ * (55/100)) * λ₁ = (ρ₂ * (45/100)) * c₂ * ΔT

Solving for ΔT

Now, we can solve the equation for ΔT:

ΔT = ((ρ₁ * (55/100)) * λ₁) / ((ρ₂ * (45/100)) * c₂)

Substituting the given values: - ρ₁ = 900 kg/m³ - ρ₂ = 1000 kg/m³ - λ₁ = 336 kJ/kg - c₂ = 4.2 kJ/(kg·°C)

We can calculate ΔT.

Calculation

Let's calculate ΔT using the given values:

ΔT = ((900 * (55/100)) * 336) / ((1000 * (45/100)) * 4.2)

Calculating this expression, we find that ΔT is approximately 8.57°C.

Final Temperature

To find the final temperature of the contents of the thermos, we need to add ΔT to the initial temperature of the boiling water. Since the initial temperature is not given, we cannot determine the final temperature without this information.

Therefore, the final temperature of the contents of the thermos after equilibrium is reached cannot be determined without knowing the initial temperature of the boiling water.

Please provide the initial temperature of the boiling water to calculate the final temperature.