For an ideal heat pump cycle COP is =

\frac{T_{H}}{T_{H} - T_{L}}

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\frac{T_{L}}{T_{H} - T_{L}}

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\frac{T_{H}}{T_{H} + T_{L}}

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\frac{T_{H}}{T_{H} \times T_{L}}

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Solution

The correct option is A $\frac{T_{H}}{T_{H} - T_{L}}$
COP (coefficient of performance $β$ ) for an heat pump cycle is defined as the ratio of heat removed per cycle $(Q_{L})$ from the surrounding to the work done per cycle ( $W$ ) to remove this heat .
$β = \frac{Q_{L}}{W}$
If $Q_{H}$ is the heat given to source (room) at temperature $T_{H}$ , then by first law of thermodynamics
$Q_{L} - Q_{H} = - W$
or $Q_{H} - Q_{L} = W$
Hence $β = \frac{Q_{L}}{Q_{H} - Q_{L}}$
or $β = \frac{(Q_{L} / Q_{H})}{(Q_{H} / Q_{H}) - (Q_{L} / Q_{H})} = \frac{Q_{L} / Q_{H}}{1 - (Q_{L} / Q_{H})}$
in a Carnot's cycle
$\frac{Q_{L}}{Q_{H}} = \frac{T_{L}}{T_{H}}$ ( $T_{L} =$ Temperature of surrounding or sink)
Therefore $β = \frac{T_{L} / T_{H}}{1 - (T_{L} / T_{H})} = \frac{T_{L}}{T_{H} - T_{L}}$

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