Question

Heat transfer of 800 kJ occurs from a source at 500 K to surroundings at 300 K. Calculate maximum available work.

• A. 320 KJ
• B. 280 KJ
• C. 420 KJ
• D. 340 KJ

Hint:

Maximum available work is equivalent to the work output of an ideal Carnot engine operating between the given temperature limits.
Just find the Carnot efficiency ($\eta_{\text{Carnot}} = 1 - T_{\text{lower}}/T_{\text{higher}}$) and multiply it by the input heat $Q$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to determine the maximum available work (also known as available energy or exergy) associated with a heat transfer process from a high-temperature source to a lower-temperature surrounding sink.

Step 2: Key Formula or Approach:
The maximum available work \(W_{\text{max}}\) is the portion of heat energy that can be completely converted into useful work by a reversible heat engine operating between the source temperature \(T\) and the surrounding temperature \(T_{0}\).
The formula is:
\[ W_{\text{max}} = Q \left( 1 - \frac{T_{0}}{T} \right) \]
where:
\(Q\) is the amount of heat transferred from the source.
\(T\) is the temperature of the heat source.
\(T_{0}\) is the temperature of the surroundings (sink).

Step 3: Detailed Explanation:

• Identify the given parameters from the question:
Heat transfer, \(Q = 800\text{ kJ}\).
Source temperature, \(T = 500\text{ K}\).
Surroundings temperature, \(T_{0} = 300\text{ K}\).

• Substitute these values into the maximum available work equation:
\[ W_{\text{max}} = 800 \left( 1 - \frac{300}{500} \right) \]

• Calculate the term inside the parentheses:
\[ 1 - \frac{3}{5} = 1 - 0.6 = 0.4 \]

• Multiply by the heat transfer value to find the maximum available work:
\[ W_{\text{max}} = 800 \times 0.4 = 320\text{ kJ} \]

• The remaining heat energy (\(800 - 320 = 480\text{ kJ}\)) represents unavailable energy that is rejected as low-grade heat to the surroundings.

Step 4: Final Answer:
The maximum available work is \(320\text{ kJ}\).