Question

Two Carnot engines operate in series between two reservoirs maintained at 600°F and 100°F, respectively. The energy rejected by the first engine is input into the second engine. If the first engine's efficiency is 20 percent greater than the second engine's efficiency, calculate the intermediate temperature.

Answer 1

The intermediate temperature is 120°F.

Step-by-step explanation:

To find the intermediate temperature, we need to compare the efficiencies of the two Carnot engines. Let's call the efficiency of the first engine 'x', and the efficiency of the second engine 'y'. Since the first engine's efficiency is 20% greater than the second engine's efficiency, we can write the equation:

x = y + 20%

Now, let's use the formula for the efficiency of a Carnot engine:

Efficiency = 1 - (Tc/Th), where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir.

The efficiency of the first engine is given by: x = 1 - (Th1/Tc1) and the efficiency of the second engine is given by: y = 1 - (Th2/Tc2).

Since the energy rejected by the first engine is input into the second engine, we can set the temperatures equal to each other: Th1 = Tc2. This means that the intermediate temperature is equal to the temperature of the hot reservoir of the second engine, Th2.

Using the equation for the first engine's efficiency, we can substitute Th1 with Tc2 and solve for Th2:

x = y + 20% ➔ 1 - (Th1/Tc1) = 1 - (Th2/Tc2) + 20%

Substituting Th1 with Tc2, we get:

1 - (Tc2/Tc1) = 1 - (Th2/Tc2) + 20%

Now, let's solve for Th2:

1 - (Tc2/Tc1) - 20% = 1 - (Th2/Tc2)

- (Tc2/Tc1) - 20% = - (Th2/Tc2)

Tc2/Tc1 + 20% = Th2/Tc2

Th2/Tc2 = Tc2/Tc1 + 20%

Th2 = Tc2 + (20%/100)*Tc2 = Tc2(1 + 20%/100)

So, the intermediate temperature is 1.2 times the temperature of the cold reservoir of the first engine, or 1.2 * 100°F = 120°F.