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An ideally efficient air conditioner keeps the room air at 290 K when the outdoor air is at 305 K. How much work does it consume when delivering 900 J of heat outside?

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opudodennis

Answer:

44.26 J

Explanation:

Efficiency=[tex]1-\frac {T_i}{T_h}[/tex] where in this case [tex]T_i[/tex] is 290 K and [tex]T_h[/tex] is 305 hence

Efficiency=[tex]1-\frac {290}{305}=0.04918[/tex]

We also know that efficiency is given by [tex]\frac {W}{Q_h}\\0.04918=\frac {W}{900}[/tex]

W=0.04918*900=44.26 J

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