Step 1: Understanding the Concept:
The efficiency of a heat engine is defined by the Second Law of Thermodynamics. For any heat engine operating between two reservoirs, the maximum theoretical efficiency is limited by the Carnot efficiency.
The Carnot efficiency depends only on the absolute temperatures of the heat source (where heat is taken in) and the heat sink (where heat is rejected).
Efficiency is a measure of how much of the input heat energy is converted into useful work.
Step 2: Key Formula or Approach:
The formula for Carnot efficiency ($\eta$) is:
\[ \eta = 1 - \frac{T_{sink}}{T_{source}} \]
where $T_{sink}$ and $T_{source}$ must be expressed in Kelvin ($K$).
Alternatively, efficiency can be viewed as $\eta = \frac{T_{source} - T_{sink}}{T_{source}}$.
Step 3: Detailed Explanation:
First, we convert all given temperatures from Celsius to the absolute Kelvin scale:
Source Temperature for all engines ($T_{source}$) $= 130 + 273.15 = 403.15 K$.
Sink Temperature for Engine A ($T_{A}$) $= 20 + 273.15 = 293.15 K$.
Sink Temperature for Engine B ($T_{B}$) $= 40 + 273.15 = 313.15 K$.
Sink Temperature for Engine C ($T_{C}$) $= 50 + 273.15 = 323.15 K$.
Now, let's analyze the efficiency formula: $\eta = 1 - \frac{T_{sink}}{T_{source}}$.
Since $T_{source}$ is constant for all three engines, the efficiency will be the highest when the ratio $\frac{T_{sink}}{T_{source}}$ is the smallest.
This ratio is minimized when $T_{sink}$ is at its lowest possible value.
Comparing the three engines:
- For A: $T_{sink} = 293.15 K$ (Lowest).
- For B: $T_{sink} = 313.15 K$.
- For C: $T_{sink} = 323.15 K$ (Highest).
Because Engine A rejects heat at the lowest temperature ($20^\circ C$), it operates across the largest temperature difference ($\Delta T$).
A larger temperature difference between the source and the sink allows for a greater portion of the thermal energy to be converted into work.
Therefore, Engine A is the most efficient.
Step 4: Final Answer:
By the Carnot principle, lower sink temperatures lead to higher efficiency for a constant source temperature. Thus, Engine A is the most efficient. The correct option is (A).