Question

В железном баке массой 5 кг находится 20 кг воды и 6 кг льда при 0 градусов. Какое количество водяного пара при 100 градусов следует ввести в бак, чтобы растопить лёд и нагреть воду до 70 градусов

Answer 1

LМп=СжМж(Т1-То)+Св(Мв+Мл)(Т1-То)+qМл.Где L-удельная теплота парообразования,q-лямбда,у меня кю-удельная теплота плавления льда.

Отсюда Mп=((Т1-То)(СжМж+Св(Мв+Мл))+qМл)/L = 4.35(кг).Ответ 4.35 кг

Answer 2

Problem Analysis

We have an iron tank with a mass of 5 kg. Inside the tank, there is 20 kg of water and 6 kg of ice at 0 degrees Celsius. We need to determine the amount of steam at 100 degrees Celsius that should be added to the tank to melt the ice and heat the water to 70 degrees Celsius.

Solution

To solve this problem, we need to consider the heat transfer that occurs during the process. The heat required to melt the ice and heat the water can be calculated using the specific heat capacities and latent heat of fusion and vaporization.

Let's break down the solution into two parts: melting the ice and heating the water.

# Melting the Ice

To melt the ice, we need to calculate the heat required to raise the temperature of the ice from 0 degrees Celsius to its melting point and then to convert it from ice to water.

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

Q1 = m1 * c1 * ΔT1

Where: - Q1 is the heat required to raise the temperature of the ice, - m1 is the mass of the ice, - c1 is the specific heat capacity of ice, - ΔT1 is the change in temperature (melting point of ice - initial temperature).

The heat required to convert the ice to water can be calculated using the formula:

Q2 = m1 * Lf

Where: - Q2 is the heat required to convert the ice to water, - m1 is the mass of the ice, - Lf is the latent heat of fusion of ice.

# Heating the Water

To heat the water, we need to calculate the heat required to raise the temperature of the water from 0 degrees Celsius to 70 degrees Celsius.

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

Q3 = m2 * c2 * ΔT2

Where: - Q3 is the heat required to raise the temperature of the water, - m2 is the mass of the water, - c2 is the specific heat capacity of water, - ΔT2 is the change in temperature (final temperature - initial temperature).

# Total Heat Required

The total heat required is the sum of the heat required to melt the ice and the heat required to heat the water:

Q_total = Q1 + Q2 + Q3

# Calculating the Amount of Steam

To calculate the amount of steam required, we need to convert the total heat required to the heat absorbed by the steam. The heat absorbed by the steam can be calculated using the formula:

Q_steam = m_steam * c_steam * ΔT_steam

Where: - Q_steam is the heat absorbed by the steam, - m_steam is the mass of the steam, - c_steam is the specific heat capacity of steam, - ΔT_steam is the change in temperature (final temperature of steam - initial temperature of steam).

Since the initial temperature of the steam is 100 degrees Celsius and the final temperature is 70 degrees Celsius, ΔT_steam = -30 degrees Celsius.

We can rearrange the formula to solve for the mass of the steam:

m_steam = Q_total / (c_steam * ΔT_steam)

Now, let's calculate the values and find the mass of steam required.

Calculation

Given: - Mass of the iron tank (m_iron) = 5 kg - Mass of water (m_water) = 20 kg - Mass of ice (m_ice) = 6 kg - Initial temperature of ice and water (T_initial) = 0 degrees Celsius - Final temperature of water (T_final) = 70 degrees Celsius - Initial temperature of steam (T_steam_initial) = 100 degrees Celsius - Final temperature of steam (T_steam_final) = 70 degrees Celsius

Specific heat capacities and latent heat values: - Specific heat capacity of ice (c_ice) = 2.09 kJ/kg°C - Specific heat capacity of water (c_water) = 4.18 kJ/kg°C - Specific heat capacity of steam (c_steam) = 2.03 kJ/kg°C - Latent heat of fusion of ice (Lf) = 334 kJ/kg

# Melting the Ice

- ΔT1 = 0 - 0 = 0 degrees Celsius - Q1 = m_ice * c_ice * ΔT1 = 6 kg * 2.09 kJ/kg°C * 0°C = 0 kJ

- Q2 = m_ice * Lf = 6 kg * 334 kJ/kg = 2004 kJ

# Heating the Water

- ΔT2 = 70 - 0 = 70 degrees Celsius - Q3 = m_water * c_water * ΔT2 = 20 kg * 4.18 kJ/kg°C * 70°C = 5836 kJ

# Total Heat Required

- Q_total = Q1 + Q2 + Q3 = 0 kJ + 2004 kJ + 5836 kJ = 7840 kJ

# Calculating the Amount of Steam

- ΔT_steam = T_steam_final - T_steam_initial = 70°C - 100°C = -30 degrees Celsius

- m_steam = Q_total / (c_steam * ΔT_steam) = 7840 kJ / (2.03 kJ/kg°C * -30°C) = -128.08 kg

Since the mass of steam cannot be negative, we can conclude that no steam needs to be added to the tank to melt the ice and heat the water to 70 degrees Celsius.

Answer

To melt the ice and heat the water to 70 degrees Celsius, no additional steam needs to be added to the tank. The heat released during the condensation of steam will be sufficient to melt the ice and heat the water to the desired temperature.