Question

4 кг воды нагревают от 288 к до кипения и получают пар при 100 градусов по цельсию .Найти массу пара,если израсходовал 1528 дж энергии

Answer 1

Масса пара = (1528 - масса воды*теплоемкость воды*(373-288)) / теплота испарения воды

Answer 2

#### Calculation of the Mass of Steam Produced To calculate the mass of steam produced when heating 4 kg of water from 28°C to its boiling point at 100°C, we need to consider the energy required for the phase change from liquid to gas. The energy required to heat a substance can be calculated using the formula: **Q = mcΔT** Where: - Q is the energy in joules (J) - m is the mass of the substance in kilograms (kg) - c is the specific heat capacity of the substance in joules per kilogram per degree Celsius (J/kg°C) - ΔT is the change in temperature in degrees Celsius (°C) In this case, we need to calculate the energy required to heat the water from 28°C to 100°C and then the energy required for the phase change from liquid to gas. 1. Energy required to heat the water from 28°C to 100°C: - The specific heat capacity of water is approximately 4.18 J/g°C or 4180 J/kg°C. - The mass of water is given as 4 kg. - The change in temperature is 100°C - 28°C = 72°C. Using the formula Q = mcΔT, we can calculate the energy required to heat the water: **Q1 = (4 kg) x (4180 J/kg°C) x (72°C)** 2. Energy required for the phase change from liquid to gas: - The heat of vaporization of water is approximately 2260 J/g or 2,260,000 J/kg. - The mass of water that undergoes the phase change is the same as the mass of steam produced. - The energy required for the phase change is given as 1528 J. Using the formula Q = mL, where L is the heat of vaporization, we can calculate the mass of steam produced: **Q2 = (m kg) x (2,260,000 J/kg) = 1528 J** Now we can solve for the mass of steam (m): **m = Q2 / (2,260,000 J/kg)** Let's calculate the mass of steam produced.