Question

Физика. идеальный тепловой двигатель получает от нагревателя количество теплоты 6МДж и отдает холодильнику количество теплоты 3МДж. вычеслите к.п.д. двигателя

Answer 1

КПД=(Q1-Q2)/Q1; КПД=(0,6*10^6-0,3*10^6))/(0,6*10^6)=0,5

Answer 2

Calculation of the Thermal Efficiency of an Ideal Heat Engine

To calculate the thermal efficiency of an ideal heat engine, we need to know the amount of heat received from the heat source (Q1) and the amount of heat rejected to the cold reservoir (Q2).

In this case, the heat engine receives 6 MJ (megajoules) of heat from the heat source (nagrevatel) and rejects 3 MJ of heat to the cold reservoir (kholodilnik).

The thermal efficiency (η) of an ideal heat engine is given by the formula:

η = (Q1 - Q2) / Q1

Substituting the given values:

η = (6 MJ - 3 MJ) / 6 MJ

Simplifying the equation:

η = 3 MJ / 6 MJ

η = 0.5

Therefore, the thermal efficiency of the heat engine is 0.5 or 50%.

Please note that the calculation assumes an ideal heat engine, which operates under certain idealized conditions and does not take into account factors such as friction, heat losses, and other inefficiencies that exist in real-world engines.