Some Carnot engine has an efficiency 0.700. The hot reservoir temperature is 900K. This means that the temperature of the cold reservoir is 270K. If the cold reservoir is dropped to 260K, the new efficiency is 0.701.
If the output of the power is kept constant by increasing the amount of time it takes to complete a cycle, what's the ratio of the wasted transfer of heat after the change to before the change?
In your first paragraph, the efficiency of a Carnot engine would be 0.711 with a 260 cold reservoir, not 0.701.
Your second paragraph does not make sense to me. Increasing the time it takes to make a cycle will not increase the power.
The ratio of wasted heat with 260 cold reservoir to that with 270 cold reservoir is (1-.711)/(1-0.700) = 289/300 = 96%
That didn't work. I think you're supposed to find work in each case but I'm not sure how to do that.
Oh wait never mind. Seems like there's some rounding issues going on with the online system. Thanks for the help drwls.
To find the ratio of the wasted transfer of heat after the change to before the change, we first need to calculate the initial and final values of wasted heat transfer.
The efficiency of a Carnot engine is given by:
Efficiency = 1 - (Tc/Th)
Where:
Efficiency is the ratio of useful work output to the heat input,
Tc is the temperature of the cold reservoir, and
Th is the temperature of the hot reservoir.
Given that the initial efficiency is 0.700, we can rearrange the equation to solve for Tc:
Tc = Th * (1 - Efficiency)
Tc = 900K * (1 - 0.700)
Tc = 900K * 0.300
Tc = 270K
From the above calculation, we can confirm that the initial temperature of the cold reservoir is 270K.
Now, with the cold reservoir dropped to 260K, we can calculate the new efficiency (Efficiency2) using the same equation with the new value of Tc:
Efficiency2 = 1 - (Tc2/Th)
Efficiency2 = 1 - (260K/900K)
Efficiency2 = 1 - 0.289
Efficiency2 = 0.711
To find the ratio of the wasted transfer of heat, we need to compare the wasted heat before (Qw1) and after (Qw2) the change. The wasted heat can be calculated using the following equation:
Qw = Qh * (1 - Efficiency)
For the initial configuration, Qw1 = Qh1 * (1 - Efficiency1):
Qw1 = Qh1 * (1 - 0.700)
For the new configuration, Qw2 = Qh2 * (1 - Efficiency2):
Qw2 = Qh2 * (1 - 0.711)
The ratio of the wasted transfer of heat after the change to before the change is then given by:
Ratio = Qw2 / Qw1
To calculate Qw1 and Qw2, we need to know the heat transfer values (Qh1 and Qh2). However, since we are keeping the output power constant by increasing the time it takes to complete a cycle, the heat transfer values remain the same. Therefore, Qw2 is proportional to Qh2, and Qw1 is proportional to Qh1.
So, the ratio of the wasted transfer of heat after the change to before the change, in this case, is equal to:
Ratio = Efficiency2 / Efficiency1
Ratio = 0.711 / 0.700
Ratio ≈ 1.016
Therefore, the ratio of the wasted transfer of heat after the change to before the change is approximately 1.016.