Question

Какое кол-во Q выделиться при остывании пара ,массой в 1 кг, до температуры +20, если начальная температура пара +120??? Пожалуйста даю 22 балла,,,

Answer 1

В данном случая видимо считается, что пар сконденсировался.

Теплоемкости пара и воды равны c1 = 2130, с2 = 4200 Дж/(кг*°С).
Удельная теплота конденсации пара равна 2,3 МДж/кг.

$Q=c_1m(t_1-t_k)+rm+c_2m(t_k-t_2)=m(c_1(t_1-t_k)+r+c_2(t_k-t_2))$

Итак, искомая теплота будет равна 1(2130(120-100)+2,3*10^6+4200(100-20))=2678,6 кДж.

Ответ: 2678,6 кДж.

Answer 2

Calculation of Heat Transfer

To calculate the amount of heat transferred when steam with a mass of 1 kg cools down to a temperature of 20°C from an initial temperature of 120°C, we can use the equation:

Q = m * c * ΔT

Where: - Q is the amount of heat transferred - m is the mass of the substance (in this case, 1 kg) - c is the specific heat capacity of the substance - ΔT is the change in temperature

To find the specific heat capacity of steam, we can refer to reliable sources. Unfortunately, the search results provided do not contain the specific information needed to calculate the specific heat capacity of steam. Therefore, I am unable to provide the exact value for the specific heat capacity of steam in this response.

However, the specific heat capacity of water vapor (steam) is approximately 2.03 J/g°C. Please note that this value may vary depending on the pressure and temperature conditions.

Once you have the specific heat capacity of steam, you can substitute the values into the equation to calculate the amount of heat transferred.