21. A heat engine operates in a cycle between temperatures 700 K and 400 K. The heat input to the engine during each cycle is 2800 J. What is the maximum possible work done by the engine in each cycle?
(A) 1200 J
(B) 1600 J
(C) 2100 J
(D) 2800 J
(E) 4400 J
I believe that the answer is E. I just don't know how to prove it. If you could show that would be great. I know that
e = |W|/|Qh|
|Qh| = |W| + |Ql|
e = 1 - |Ql|/|Qh|
wait no the answer is A how though?
In a maximum-efficiency "Carnot" cycle, the ratio
Qout/Qin = Tout/Tin
because the entropy change when heat is absorbed and released by the fluid is the same. T remains the same during each process.
Therefore e = 1 - Tout/Tin = 3/7, and
Wout = 3/7 * 2800 = 1200 J
Practical engines do not use a Carnot cycle, but come reasonably close.
To find the maximum possible work done by the engine in each cycle, we can use the Carnot efficiency formula.
The Carnot efficiency (e) is given by:
e = 1 - (|Ql| / |Qh|)
where |Qh| is the heat input to the engine during each cycle, and |Ql| is the heat rejected by the engine during each cycle.
In this case, |Qh| = 2800 J (given).
To find |Ql|, we can use the formula:
|Ql| = |Qh| - |W|
where |W| is the work done by the engine during each cycle.
Substituting the values:
|Ql| = 2800 J - |W|
Now, we can substitute this value of |Ql| in the Carnot efficiency formula:
e = 1 - (|Ql| / |Qh|)
= 1 - ((2800 J - |W|) / 2800 J)
To maximize the work done by the engine, we want to find the maximum value of e. This occurs when the fraction ((2800 J - |W|) / 2800 J) is minimized.
To minimize this fraction, the numerator (2800 J - |W|) should be minimized. This occurs when |W| is maximized.
Since |W| represents the work done by the engine, the maximum possible work done by the engine is when |W| = 2800 J.
Therefore, the maximum possible work done by the engine in each cycle is 2800 J.
So, the correct answer is (D) 2800 J.
To find the maximum possible work done by the engine in each cycle, we can use the formula for efficiency:
efficiency (e) = work output (W) / heat input (Qh)
Given that the heat input (Qh) is 2800 J, we need to determine the heat output (Ql) in order to find the work output (W).
Now, let's consider the Carnot Cycle, which is an idealized heat engine operating between two temperatures.
In the Carnot Cycle, the efficiency can be expressed as:
efficiency (e) = 1 - (Ql / Qh)
Where Ql is the heat output.
We are given the temperatures of the engine, 700 K (Thigh) and 400 K (Tlow).
Based on the Carnot Cycle, the efficiency can also be expressed as the ratio of the temperature differences:
efficiency (e) = (Thigh - Tlow) / Thigh
Let's substitute the given temperatures into the equation:
e = (700 K - 400 K) / 700 K
e = 300 K / 700 K
Now, we can calculate the value of efficiency (e):
e ≈ 0.4286
To find the work output (W), we can rearrange the efficiency formula:
W = e * Qh
W = 0.4286 * 2800 J
W ≈ 1200 J
Therefore, the maximum possible work done by the engine in each cycle is approximately 1200 J.
So, the correct answer is (A) 1200 J.