Question

В сосуде, расположенном в термостате, находится 2 кг воды при температуре 50 °С. Под слой жидкости подаётся пар таким образом, что он полностью конденсируется. Температура пара 100 °С. Определите, какое количество пара (в граммах) было подано под слой жидкости, если температура жидкости в конце процесса теплообмена стала равна 75 °С. Удельная теплота парообразования равна 2300 кДж/кг, удельная теплоёмкость воды 4200 Дж/(кг · °С). Ответ округлите до целого числа.

Answer 1

Ответ:

Объяснение: m1=?

Кол-во теплоты,необходимое для нагревания воды

Q1=Св*m*(75-50)

Кол-во теплоты необходимое для перехода пара из насыщенного состояния в водяной пар.

Q2=-(q*m1) где q-удельная теплота параобразования

Кол-во теплоты,необходимое для остужения пара

Q3=Cв*m1(75-100)

Q1+Q2+Q3=0

приравниваешь и получаешь ответ.Я не уверен,что 3 ур-е нужно,проверь на всякий случай

Answer 2

Problem Analysis

We are given a vessel in a thermostat containing 2 kg of water at a temperature of 50 °C. Steam is supplied under the liquid layer in such a way that it completely condenses. The temperature of the steam is 100 °C. We need to determine the amount of steam (in grams) supplied under the liquid layer if the temperature of the liquid at the end of the heat exchange process becomes 75 °C. We are also given the specific heat of vaporization of water (2300 kJ/kg) and the specific heat capacity of water (4200 J/(kg·°C)).

Solution

To solve this problem, we can use the principle of energy conservation. The heat gained by the water during the process will be equal to the heat lost by the steam. We can calculate the heat gained by the water using the formula:

Q = m * c * ΔT

Where: - Q is the heat gained by the water - m is the mass of the water - c is the specific heat capacity of water - ΔT is the change in temperature of the water

We can calculate the heat lost by the steam using the formula:

Q = m * L

Where: - Q is the heat lost by the steam - m is the mass of the steam - L is the specific heat of vaporization of water

Since the steam completely condenses, the mass of the steam will be equal to the mass of the water at the end of the process. We can calculate the mass of the water at the end of the process using the formula:

m = m_initial + m_steam

Where: - m is the mass of the water at the end of the process - m_initial is the initial mass of the water - m_steam is the mass of the steam supplied under the liquid layer

By equating the heat gained by the water and the heat lost by the steam, we can solve for the mass of the steam:

m_steam = (m * c * ΔT) / L

Now we can substitute the given values into the equation and calculate the mass of the steam.

Calculation

Given: - m_initial = 2 kg (mass of water at the beginning) - T_initial = 50 °C (temperature of water at the beginning) - T_final = 75 °C (temperature of water at the end) - T_steam = 100 °C (temperature of the steam) - L = 2300 kJ/kg (specific heat of vaporization of water) - c = 4200 J/(kg·°C) (specific heat capacity of water)

First, we need to convert the specific heat of vaporization from kJ/kg to J/kg: L = 2300 kJ/kg * 1000 J/kJ = 2,300,000 J/kg

Next, we can calculate the change in temperature of the water: ΔT = T_final - T_initial = 75 °C - 50 °C = 25 °C

Now we can substitute the values into the equation to calculate the mass of the steam: m_steam = (m * c * ΔT) / L = (2 kg * 4200 J/(kg·°C) * 25 °C) / 2,300,000 J/kg

Calculating the mass of the steam: m_steam = (2 * 4200 * 25) / 2300000 = 0.0913 kg

Finally, we need to convert the mass of the steam from kg to grams: m_steam = 0.0913 kg * 1000 g/kg = 91.3 g

Answer

The amount of steam supplied under the liquid layer is approximately 91 grams.