I think I also forgot to type out another step I took. Using the efficiency equation, I did 1787 TWh/.4 and got 4467.5 TWh.
I used 4467.5 TWh and 1787 TWh to find the waste heat energy. So I did 4467.5-1787 and got 2680.5 TWh. Because this is still in TWh, I converted it to Joules. (Should I...
Homework Statement
In the year 2004 the USA produced 1787 TWh of electrical energy in conventional thermal plants and 476 TWh in nuclear plants. Assuming 30% efficiency for nuclear plants and 40% for conventional thermal plants, determine the (annual) volume of cooling water required to cool...
Homework Statement
The door of a refrigerator is 1.5 m high, 0.80 m wide and 6.0 cm thick. Its thermal conductivity is 0.21Wm-1 degrees Celsius-1.
a) what is the heat loss per hour through the door neglecting convection effects?
b) Air is usually still inside the refrigerator so there will be a...
Right, but if the heat is coming from the hot reservoir and the engine is absorbing the 500 J from the hot reservoir, then doesn't that still make it sound like the 500 J is Qc?
So in this case, I shouldn't have taken the hot reservoir into account? Only the engine and the atmosphere? So the atmosphere was the "cold reservoir" and the engine was the hot reservoir?
No, I still don't get it. I never said that Qh and Qc were equal from the start. I always knew that. My problem is, 500 J was *absorbed*, but if something is absorbed by something else (since heat only travels from high to low temperature) that means it had to have been to a cold body from a hot...
No, it's not equal. There's the hot reservoir, and then the engine (which is the part that does work), and then the cold reservoir. So heat is ejected from the hot reservoir (Qh) to the engine. The engine uses some of the heat to do work and then ejects the rest of the heat to the cold reservoir...
Qh is the amount of heat rejected by the hot reservoir. Qc is the amount of heat absorbed by cold reservoir from hot reservoir. This is how I learned it.
But if heat is taken from the hot reservoir by another body, then wouldn't that be Qc?
I posted the equations in the original question.
Efficiency=1-(Qc/Qh)
So .2=1-(500/Qh) ( This is what I did.)
So from that equation I got Qh=625 J.
And if work done=Qh-Qc then 625-500 is 125 J of work done.
Because it says the heat 'absorbed' is 500 J and heat can only go from high to low, right? So if I thought the given 500 J was Qc then I got 625 solving for Qh and then the work done was 125.
Well the reason i didn't put what my answer was is because I want to see what you think the question is asking. Let's just say that I thought 500 was Qc. From there, I solved the problem correctly if 500 was Qc.
I had an exam last week and I just got it back today. On the exam was a question that I got wrong even though his wording was terrible (he's from India) and I feel that it was not clearly expressed what he was saying. The question is: "A heat engine is 20% efficient. If it absorbs 500 J of heat...
Yes. So it has the same velocity as Earth. But I'm still confused as to how to find the velocity of earth. Even if I use the equation 2pir/T, I get .465 m/s. And when I plug that into the equation GM/r=v^2, I get a very large number that must be incorect.