A Carnot engine, operating between hot and cold reservoirs whose temperatures are 800.0 and 400.0 K, respectively, performs a certain amount of work W1. The temperature of the cold reservoir is then lowered to 200.0 K (see the drawing), and the work done by the engine is W2. Assuming that the input heat QH is the same in both cases, what is the ratio W2/W1?
The W2/W1 ratio equals the efficiency ratio, since the heat input is constant.
Initially, the Carnot efficiency is
1 - (Tc/Th) = 1 - 400/800 = 0.50
After lowering Tc, the efficiency becomes
1 - (Tc/Th) = 1 - 200/800 = 0.75
W2/W1 = 0.75/0.50 = 1.5
To find the ratio W2/W1, we need to understand the principles of a Carnot engine and use the equation that relates work done by the engine to the temperatures of the hot and cold reservoirs.
A Carnot engine is a theoretical heat engine that operates on the reversible Carnot cycle. It consists of four steps: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression.
In this scenario, the Carnot engine operates between two temperature reservoirs: a hot reservoir at 800.0 K and a cold reservoir initially at 400.0 K. The engine performs work W1.
To calculate W1, we can use the equation:
W1 = QH - QC,
where QH is the input heat from the hot reservoir, and QC is the heat released to the cold reservoir.
Given that the Carnot engine is operating between two isothermal reservoirs, the efficiency (η) of the Carnot engine can be expressed as:
η = 1 - (TC / TH),
where TC is the temperature of the cold reservoir and TH is the temperature of the hot reservoir.
Rearranging this equation, we can solve for the heat released to the cold reservoir (QC):
QC = TH - TC.
Substituting this back into the equation for work (W1), we get:
W1 = QH - QC = QH - (TH - TC).
Now, with the temperature of the cold reservoir lowered to 200.0 K, we need to find the new work done by the engine, W2.
Using the same equation, but with the new temperature for the cold reservoir, we have:
W2 = QH - QC' = QH - (TH - TC'),
where QC' is the heat released to the cold reservoir at the new temperature TC'.
Now, we can find the ratio W2/W1:
(W2 / W1) = (QH - (TH - TC')) / (QH - (TH - TC)).
Since the input heat, QH, is the same in both cases, it cancels out:
(W2 / W1) = (TH - TC') / (TH - TC).
Substituting the known values:
(W2 / W1) = (800.0 K - 200.0 K) / (800.0 K - 400.0 K).
Simplifying further:
(W2 / W1) = 600.0 K / 400.0 K.
Finally, performing the division:
(W2 / W1) = 1.5.
Therefore, the ratio W2/W1 is 1.5.
To find the ratio of \( \frac{W_2}{W_1}\), we can use the formula for the efficiency of a Carnot engine. The efficiency (\( \eta \)) of a Carnot engine is given by:
\[ \eta = 1 - \frac{T_c}{T_h} \]
Where \( T_c \) is the temperature of the cold reservoir and \( T_h \) is the temperature of the hot reservoir.
Since the input heat \( Q_H \) is the same in both cases, we can write:
\[ W_1 = Q_H(1 - \eta) \]
\[ W_2 = Q_H(1 - \eta') \]
Where \( \eta' \) is the new efficiency of the Carnot engine with the lower temperature of the cold reservoir.
To find \( \eta' \), we need the new temperatures of the hot and cold reservoirs. The hot reservoir temperature remains the same at 800.0 K, but the cold reservoir temperature decreases to 200.0 K.
Substituting the values into the efficiency equation, we can calculate the ratio \( \frac{W_2}{W_1} \):
\[ \eta = 1 - \frac{400.0}{800.0} = 0.5 \]
\[ \eta' = 1 - \frac{200.0}{800.0} = 0.75 \]
\[ \frac{W_2}{W_1} = \frac{Q_H(1 - \eta')}{Q_H(1 - \eta)} = \frac{1 - \eta'}{1 - \eta} \]
\[ \frac{W_2}{W_1} = \frac{1 - 0.75}{1 - 0.5} = \frac{0.25}{0.5} = 0.5 \]
Therefore, the ratio of \( \frac{W_2}{W_1} \) is 0.5.