Question

Теплова машина за цикл роботи одержує від нагрівника кількість теплоти, яка дорівнює 100 Дж, і віддає холодильнику кількість теплоти, що становить 40 Дж. Яким є ККД теплової машини?

Answer 1

Q1= 100 Q2= 40 η-?

η = Q1-Q2/Q1 *100%= 100-40/100 *100% = 60

Answer 2

Calculating the efficiency of a heat engine

The question asks about calculating the efficiency of a heat engine based on the amounts of heat absorbed from a hot reservoir (Qh) and rejected to a cold reservoir (Qc) during one full cycle.

The heat engine absorbs 100 Joules (Qh) from the hot reservoir and rejects 40 Joules (Qc) to the cold reservoir.

The efficiency (η) of a heat engine is defined as the ratio of useful work done (W) to the heat absorbed from the hot reservoir (Qh):

`η = W / Qh`

We can calculate the useful work done (W) using the First Law of Thermodynamics:

`W = Qh - Qc`

Plugging in the values given:

`W = Qh - Qc = 100 J - 40 J = 60 J`

Then the efficiency is:

`η = W / Qh = 60 J / 100 J = 0.6 = 60%`

Therefore, the efficiency of the heat engine is 60%.