Heat Q flows spontaneously from a reservoir at

[copitlory8]

Heat Q flows spontaneously from a reservoir at 383 K into a reservoir that has a lower temperature T. Because of the spontaneous flow, thirty percent of Q is rendered unavailable for work when a Carnot engine operates between the reservoir at temperature T and a reservoir at 234 K. Find the temperature T.

The initial temperature is 383K. I have no idea what this problem is about because my teacher didnot cover the material. Please help me with the solution.

 

[Chi Meson]

You are using the Cutnell and Johnson text aren't you? I used to teach with that and I remember this problem specifically stumped nearly all of my students. I am still not 100% certain as to what the guy meant to say when he wrote the problem. Here's my interpretation:

Look in the text for "Carnot Efficiency." That would be the efficiency of the best possible heat engine, one that had no friction at all. It's an efficiency based on temperatures of the two reservoirs. Compare Carnot efficiency to regular heat engine efficiency (write them out side by side), and notice what ratios are equal to each other

The unknown temperature T is in between 383K and 284K. Plug these into the carnot efficiency formula. You can get the efficiency here, but you won't need it.

OK, "unfortunately there was a spontaneous flow of heat to an intermediate reservoir, before it got to the machine, lowering Qhot by 30 %."

Go back to the comparison of the two efficiency formulas. Look at the two ratios that are similar. If Qhot is 30% less, what else ...

I just lead you to the front door. Go in ...

 

[copitlory8]

umm. i not really sure where to go