a heat engine with an efficiency of 30.0% performs 2500 j of work. how much heat is discharged to the lower temperature reservoir

Q₂ = 5833.33 J

Explanation:

First we need to find the energy supplied to the heat engine. The formula for the efficiency of the heat engine is given as:

η = W/Q₁

where,

η = efficiency of engine = 30% = 0.3

W = Work done by engine = 2500 J

Q₁ = Heat supplied to the engine = ?

Therefore,

0.3 = 2500 J/Q₁

Q₁ = 2500 J/0.3

Q₁ = 8333.33 J

Now, we find the heat discharged to lower temperature reservoir by using the formula of work:

W = Q₁ - Q₂

Q₂ = Q₁ - W

where,

Q₂ = Heat discharged to the lower temperature reservoir = ?

Therefore,

Q₂ = 8333.33 J - 2500 J

Q₂ = 5833.33 J

the inner core.

good luck: )

klicker