Calculate Heat Rejected by 200W Lamp in 1 Hour

AI Thread Summary
A 200W lamp converts to 682.4 BTU/hr when powered for one hour. With 700 BTU extracted from the heat source and the engine operating at 50% efficiency, only 350 BTU can be used for work. The heat rejected by the system is calculated to be 17.6 BTU. Given the energy consumption, the lamp can be powered for less than 2 hours with the available energy. The calculations illustrate the relationship between energy input, efficiency, and heat rejection in a heat engine system.
Windseaker
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Given :
QH=W+QC and Power= E/t or W/t

A heat engine is attached to a 200W lamp. If we power the lamp for 1 hour and extract 700 btu from the heat source, how much heat is rejected?

Attempt:

200w(3.412btu/1w) = 682.4 btu now, 700 btu-682.4 btu =17.6 btu rejected??
 
Last edited:
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Yes.
 
what if i was after time?Given :
QH=W+QC and Power= E/t or W/t

A heat engine is attached to a 200W lamp. we can extract 700 btu from the heat source and engine is 50% efficient, how long can we power it??

Attempt:

200w(3.412btu/1w) =682.4 /2 btu =341.2 btu and 700 btu-682.4 btu =17.6 btu
Total 358.8btu
lost from there??
 
Last edited:
Windseaker said:
what if i was after time?


Given :
QH=W+QC and Power= E/t or W/t

A heat engine is attached to a 200W lamp. we can extract 700 btu from the heat source and engine is 50% efficient, how long can we power it??

Attempt:

200w(3.412btu/1w) =682.4 /2 btu =341.2 btu and 700 btu-682.4 btu =17.6 btu
Total 358.8btu
lost from there??
You are converting watts (Joules/sec) into BTU/hr (1 w = 3.412 btu/hr). Therefore, the 200w lamp consumes 682.4 btu/hr.

If the Qh is 700 btu at 50% efficiency, this means that you can extract 350 btu of work. Since the light consumes 682.4 btu in one hour, how long can you keep the light going with 350 btu of energy?

AM
 
less then 2hrs.
 
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