Consider an inefficient engine which draws 200 J of heat from a hot reservoir at 100oC, converts 48 J to work, and rejects 152 J to a cold reservoir at 0oC. Theoretically, how much maximum work can be produced if the efficiency of the engine is maximized? How much work was “lost” by using this inefficient heat engine design? (Hint: Carnot Engine)

No cheating

e maximum = 27%

To determine the maximum work and the amount of work lost by using an inefficient heat engine, we can apply the principles of the Carnot engine.

The efficiency (η) of any heat engine can be calculated using the formula:

η = 1 - (Qc / Qh)

Where Qc is the amount of heat rejected to the cold reservoir, and Qh is the amount of heat drawn from the hot reservoir.

In this case, Qc = 152 J and Qh = 200 J.

So, the efficiency can be calculated as:

η = 1 - (152 / 200)
η = 1 - 0.76
η = 0.24 or 24%

Now, the efficiency of the Carnot engine is given by the formula:

η_carnot = 1 - (Tc / Th)

Where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir.

In this case, Tc = 0°C + 273.15 = 273.15 K and Th = 100°C + 273.15 = 373.15 K.

So, the efficiency of the Carnot engine can be calculated as:

η_carnot = 1 - (273.15 / 373.15)
η_carnot = 0.266 or 26.6%

The maximum work that can be produced is given by the formula:

W_max = η_carnot * Qh

Substituting the values, we get:

W_max = 0.266 * 200
W_max = 53.2 J

Therefore, the maximum work that can be produced, if the efficiency is maximized, is 53.2 J.

To find the amount of work lost, we can subtract the actual work done (48 J) from the maximum work:

Amount of work lost = W_max - Actual work done
Amount of work lost = 53.2 J - 48 J
Amount of work lost = 5.2 J

Hence, the work lost by using this inefficient heat engine design is 5.2 J.