Question

В идеальной тепловой машине за счёт каждого килоджоуля энергии, получаемого от нагревателя, свершается работа 400 Дж. определите КПД машины и температуру нагревателя (в градусах по шкалам Кельвина и Цельсия), если температура холодильника 27 °C.

Answer 1

Кпд равен 40%
(Qн -Qх)/Qн= КПД
Следовательно:
(Qн - 300)/Qн= 0,4
Решаем, получается 500 по кельвину или 227 по цельсия

Answer 2

Calculation of Efficiency and Temperature of the Heat Source

To determine the efficiency (η) and temperature of the heat source in an ideal heat engine, we can use the formula for efficiency:

η = (Work output / Heat input) x 100%

Given that the work output is 400 J and the heat input is the energy obtained from the heat source, we can rewrite the formula as:

η = (400 J / Q) x 100%

where Q is the energy obtained from the heat source.

To find the temperature of the heat source, we need to use the relationship between temperature and heat energy:

Q = mcΔT

where Q is the heat energy, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

Since we don't have information about the mass or specific heat capacity, we cannot calculate the exact temperature of the heat source. However, we can still determine the temperature difference between the heat source and the cold reservoir.

Given that the temperature of the cold reservoir (the refrigerator) is 27 °C, we can convert it to Kelvin by adding 273.15:

T_cold = 27 °C + 273.15 = 300.15 K

To find the temperature of the heat source, we need to know the temperature difference (ΔT) between the heat source and the cold reservoir. Unfortunately, we don't have enough information to calculate the exact temperature of the heat source.

Therefore, we can conclude that the efficiency (η) and the temperature of the heat source cannot be determined with the given information.

Please provide additional information if available, such as the mass or specific heat capacity, to calculate the temperature of the heat source and the efficiency of the machine accurately.