Question

В сосуд, имеющий температуру 0°С, впустили пар массой 1 кг при температуре 100°С. Сколько воды изначально было в сосуде, если через некоторое время в нём установилась температура 20°С? Удельная теплоёмкость воды равна 4200 Дж/(кг·°С), удельная теплота парообразования воды - 2,3 МДж/кг.

Answer 1

MвcвΔtв=mпL+mпcвΔtвп
mв=(mпL+mпcвΔtвп)/(cвΔtв)
mв=(1•2300000+1•4200•80)/(4200•20)=31,4 кг

Answer 2

Problem Analysis

We are given a vessel with an initial temperature of 0°C. We introduce 1 kg of steam at a temperature of 100°C into the vessel. After some time, the temperature in the vessel reaches 20°C. We need to determine the initial amount of water in the vessel.

Solution

To solve this problem, we can use the principle of conservation of energy. The energy gained by the water in the vessel is equal to the energy lost by the steam.

Let's denote the initial amount of water in the vessel as m (in kg). The specific heat capacity of water is given as 4200 J/(kg·°C). The specific latent heat of vaporization of water is given as 2.3 MJ/kg.

The energy gained by the water can be calculated using the formula:

Energy gained by water = m × specific heat capacity of water × change in temperature

The energy lost by the steam can be calculated using the formula:

Energy lost by steam = mass of steam × specific latent heat of vaporization of water

Since the energy gained by the water is equal to the energy lost by the steam, we can set up the following equation:

m × specific heat capacity of water × change in temperature = mass of steam × specific latent heat of vaporization of water

We can rearrange this equation to solve for m:

m = (mass of steam × specific latent heat of vaporization of water) / (specific heat capacity of water × change in temperature)

Let's substitute the given values into this equation and calculate the initial amount of water in the vessel.

Calculation

Given: - Mass of steam = 1 kg - Specific heat capacity of water = 4200 J/(kg·°C) - Specific latent heat of vaporization of water = 2.3 MJ/kg = 2.3 × 10^6 J/kg - Change in temperature = 20°C - 0°C = 20°C

Substituting these values into the equation:

m = (1 kg × 2.3 × 10^6 J/kg) / (4200 J/(kg·°C) × 20°C)

Simplifying the equation:

m = (1 × 2.3 × 10^6) / (4200 × 20) kg

Calculating the value of m:

m ≈ 27.38 kg

Therefore, the initial amount of water in the vessel was approximately 27.38 kg.

Answer

The initial amount of water in the vessel, if the temperature in the vessel eventually reaches 20°C after introducing 1 kg of steam at 100°C, is approximately 27.38 kg.