A house loses energy through the exterior walls and roof at...

A house loses energy through the exterior walls and roof at a rate of $5000 \mathrm{~J} / \mathrm{s}=5.00 \mathrm{~kW}$ when the interior temperature is $22.0^{\circ} \mathrm{C}$ and the outside temperature is $-5.00^{\circ} \mathrm{C}$. Calculate the electric power required to maintain the interior temperature at $22.0^{\circ} \mathrm{C}$ for the following two cases: (a) The electric power is used in electric resistance heaters (which convert all of the electricity supplied into internal energy). (b) The electric power is used to drive an electric motor that operates the compressor of a heat pump (which has a coefficient of performance [COP] equal to $60.0 \%$ of the Carnot-cycle value).

A house loses energy through the exterior walls and roof at a rate of $5000 \mathrm{~J} / \mathrm{s}=5.00 \mathrm{~kW}$ when the interior temperature is $22.0^{\circ} \mathrm{C}$ and the outside temperature is $-5.00^{\circ} \mathrm{C}$. Calculate the electric power required to maintain the interior temperature at $22.0^{\circ} \mathrm{C}$ for the following two cases:
(a) The electric power is used in electric resistance heaters (which convert all of the electricity supplied into internal energy). (b) The electric power is used to drive an electric motor that operates the compressor of a heat pump (which has a coefficient of performance [COP] equal to $60.0 \%$ of the Carnot-cycle value).

Key Concepts

Carnot Cycle Efficiency

The Carnot cycle describes an idealized thermodynamic process that provides the maximum possible efficiency—that is, the highest COP—for heat engines or heat pumps operating between two thermal reservoirs. Real-world devices often achieve only a fraction of the Carnot efficiency, and understanding this limit helps in comparing and improving practical thermal systems.

Coefficient of Performance (COP) of a Heat Pump

The coefficient of performance (COP) measures the effectiveness of a heat pump, defined as the ratio of heat output to the work input needed to accomplish that transfer. It indicates how much heat is delivered per unit of electrical energy consumed, and is a key factor in determining the efficiency of a heat pump compared to direct electric heating.

Electrical Resistance Heating

Electrical resistance heating involves converting electric energy directly into thermal energy with nearly 100% efficiency. In this method, the electrical power supplied equates almost directly to the heat energy produced, making its energy management straightforward.

Thermal Energy Loss

Thermal energy loss is the rate at which energy is transferred out of a system due to a temperature difference between the interior and the exterior. It’s measured in watts (joules per second) and represents how much heating energy must be supplied to maintain a steady internal temperature.

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