Question

Question: Suppose a heat pump has a stationary bicycle attachment that allows you to provide the work instead of using an electrical wall outlet. If your heat pump has a coefficient of performance of 2.0 and you can cycle at a racing pace (Table 15–2) for a half hour, how much heat can you provide?

Short answer

The heat delivered to the heat pump is \(4.572 \times {10^6}\;{\rm{J}}\).

Step-by-step solution

Determination of the coefficient of performance

The coefficient of performance of a heat pump can be calculated by dividing the heat intake by the pump by the work performed by the pump.

Given information

The coefficient of performance of the pump is \({\rm{COP}} = {\rm{2}}{\rm{.0}}\).

Evaluation of the work input to the heat pump

From table 15-2, the metabolic rate for bicycling is \(1270\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {\rm{s}}}} \right.} {\rm{s}}}\).

The work input to the heat pump is calculated below:

\(\begin{aligned}{l}W &= \left( {1270\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {\rm{s}}}} \right.} {\rm{s}}}} \right)\left( {0.5\;{\rm{h}}} \right)\left( {\frac{{3600\;{\rm{s}}}}{{1\;{\rm{h}}}}} \right)\\W &= 2.286 \times {10^6}\;{\rm{J}}\end{aligned}\)

Evaluation of the heat delivered to the heat pump

The heat delivered to the heat pump is calculated below:

\(\begin{aligned}{l}{Q_{\rm{H}}} &= W\left( {{\rm{COP}}} \right)\\{Q_{\rm{H}}} &= \left( {2.286 \times {{10}^6}\;{\rm{J}}} \right)\left( {2.0} \right)\\{Q_{\rm{H}}} &= 4.572 \times {10^6}\;{\rm{J}}\end{aligned}\)

Thus, the heat delivered to the heat pump is \(4.572 \times {10^6}\;{\rm{J}}\).