A theoretical gasoline engine runs with 76 percent efficiency. It is calculated that it expels 3.25*10^4 J of heat.

a. Find the heat absorbed in one cycle.
b. Find the work output in one cycle.

All this comes from the definition of efficiency.

percentefficiency*100= work out/ heat in

and of course, work out= heat in - expelled energy.

To solve this problem, we can follow the given formula for efficiency:

efficiency = (work output / heat input) * 100

We are given that the efficiency of the gasoline engine is 76 percent. Let's represent the heat input as H and the work output as W.

a. Find the heat absorbed in one cycle:
Since the heat expelled is given as 3.25 * 10^4 J, we can use the formula mentioned earlier:

76% = (W / H) * 100

Dividing both sides by 100, we get:

0.76 = W / H

Rearranging the equation, we find:

W = 0.76 * H

Substituting the given expelled energy value (3.25 * 10^4 J), we can solve for H:

W = 0.76 * H
3.25 * 10^4 J = 0.76 * H

Dividing both sides of the equation by 0.76, we get:

H = (3.25 * 10^4 J) / 0.76

Calculating this value yields the heat absorbed in one cycle.

b. Find the work output in one cycle:
From the formula mentioned earlier, we know that:

W = H - expelled energy

Substituting the calculated heat input (H) and the given expelled energy (3.25 * 10^4 J), we can find the work output:

W = (3.25 * 10^4 J) - (3.25 * 10^4 J)

Calculating this value gives us the work output in one cycle.

By following these steps and applying the given formulas, you can find both the heat absorbed and the work output in one cycle of the theoretical gasoline engine.