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An engine does 18500 J of work and rejects 7810 J of heat into a cold reservoir whose temperature is 285 K. What would be the smallest possible temperature of the hot reservoir

Sagot :

hamzaahmeds

This question involves the concepts of thermal efficiency and hot reservoir.

The smallest possible temperature of the hot reservoir would be "960 k".

The thermal efficiency of an engine is given by the following formula:

[tex]Efficiency=\frac{Work}{Heat\ In}=\frac{Work}{Work+Heat\ Rejected}\\\\Efficiency = \frac{18500\ J}{18500\ J+7810\ J}\\\\Efficiency = 0.703[/tex]

Another formula for thermal efficiency of an engine is:

[tex]Efficiency = 1-\frac{T_2}{T_1}[/tex]

where,

T₁ = Temperature of the hot reservoir = ?

T₂ = Temperature of the cold reservoir = 285 k

Therefore,

[tex]0.703=1-\frac{285\ k}{T_1}\\\\\frac{285\ k}{T_1}=1-0.703\\\\T_1=\frac{285\ k}{0.297}[/tex]

T₁ = 960 k

Learn more about thermal efficiency here:

https://brainly.com/question/1332182?referrer=searchResults

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