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 5 000 J/s ! 5.00 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 energy transferred in by electrical transmission into internal energy). (b) What If ? The electric power is used to drive an electric motor that operates the compressor of a heat pump, which has a coefficient of performance equal to 60.0% of the Carnot-cycle value.

Key Concepts

Carnot Cycle

The Carnot cycle represents an idealized thermodynamic cycle that provides an upper limit on the efficiency that any heat engine or heat pump can achieve between two thermal reservoirs. By comparing the actual performance of a heat pump to that of the Carnot cycle, engineers can assess the extent to which a system approaches theoretical efficiency limits. This concept is crucial when evaluating the performance of heat pumps and understanding how factors like temperature differences contribute to overall device efficiency.

Heat Pumps

Heat pumps are devices that move heat from a cooler area to a warmer area using mechanical work, typically via a refrigeration cycle. They are used not only for heating but also for cooling. Heat pumps often achieve higher overall efficiency than direct electric heating by leveraging a refrigeration cycle, where a small amount of work can move a much larger quantity of heat, as indicated by their coefficient of performance.

Coefficient of Performance (COP)

The coefficient of performance (COP) is a measure of the efficiency of heating and cooling devices, such as heat pumps. It is defined as the ratio of useful heat transfer to the input work required. A higher COP indicates a more efficient system as it delivers more heat per unit of energy consumed. When operating below the ideal Carnot-cycle maximum, real-world heat pumps achieve a COP that is a fraction of the maximum possible value, which has significant implications for the electric power required for heating.

Heat Transfer

Heat transfer is the process through which thermal energy is exchanged between systems or their surroundings, often occurring through conduction, convection, or radiation. In the context of building energy loss, heat transfer principles help us understand how energy escapes from a structure through its walls, roof, and other surfaces due to temperature gradients, and how this affects the rate at which energy must be added to maintain a desired interior temperature.

Electric Resistance Heating

Electric resistance heating refers to the conversion of electrical energy directly into thermal energy through the resistance of a material. This process is nearly 100% efficient at converting electricity into heat, meaning that the amount of electrical power consumed is directly equivalent to the thermal energy produced. It is a straightforward method of heating in which the energy input exactly matches the amount of energy lost through the structure.

Transcript

00:03 Here we are given that the walls of room loses energy at a rate of 5 ,000 joules per second that is equal to 5 kilowatt and we want to keep the temperature of room constant at 22 degrees celsius.

00:31 So here the first method is to run an electric heater and the electric heater converts all the electric energy into thermal energy so in order to keep the temperature of room constant energy loss from the room must be equal to energy edit to the room by electric heater as electric heater converts all the electric energy into thermal energy so we need electric power for electric heater equal to 5 kilowatt so that is power for heater equal to 5 kilowatt now in the second part we want to keep the temperature of room constant by using heat pump and we want to find the amount of energy needed to run the motor of compressor of heat pump first we will calculate here coefficient of performance for ideal heat pump so here we will write first coefficient of performance for ideal heat pump is equal to heat deposit at the hot reservoir that is the desired output divided by the input that is the work done on the heat pump which is actually equal to the heat deposit at hard reservoir minus heat extracted from coal reservoir that is also equal to temperature of hot reservoir divided by temperature of hot reservoir minus temperature of coal reservoir as temperature of hard reservoir that is the temperature of interior of room which is 22 degrees celsius that is equal to 295 kelvin divided by 295 kelvin and temperature outside the room is minus 5 degrees celsius that is equal to 260 a calvian this gives us the coefficient of performance for ideal heat pump working on the temperatures equal to 10 .9 to 6 now we are given that the heat pump working in the room has coefficient of performance equal to 60 % of the ideal heat pump.

04:18 So we can write here coefficient of performance of the heat pump used in the room is equal to 60 % of the coefficient of performance of ideal heat pump...

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