Energy problem dealing with heat extracted from cold air

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Homework Help Overview

The problem involves a heat pump that requires a specific amount of electrical power to deliver heat to a house, and it asks for the amount of energy extracted from the cold air outside per second. The subject area is thermodynamics, specifically focusing on heat transfer and the operation of heat pumps.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to use the coefficient of performance (COP) equation but indicates that their application was incorrect. Some participants question the fate of the electrical energy used by the pump, while others draw parallels to the operation of an air conditioner and discuss the thermodynamic principles involved.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. There is a focus on understanding the relationship between the heat extracted and the work done by the heat pump, as well as the implications of reversible processes in thermodynamics.

Contextual Notes

Participants are working within the constraints of the problem statement and are questioning the assumptions related to the operation of the heat pump and the definitions of the variables involved.

astru025
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Homework Statement



A heat pump requires 385 W of electrical power to deliver heat to your house at a rate of 2410 J per second. How many joules of energy are extracted from the cold air outside each second?


Homework Equations



COP= Qc / W . This was the only equation I could find.

The Attempt at a Solution


Any help would be nice! I tried the above equation but I turned out to be incorrect.
 
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use this and find out the ans...
 
What do you think happens to the electrical energy used by the pump?
 
This is the same as an air conditioner that is cooling the outside and rejecting the heat to the inside. If the heat pump operates reversibly, then Q1 is removed from the outside at temperature T1, and Q2 is added to the inside at temperature T2. For a reversible process, such that ΔS of the surroundings = 0, \frac{Q_1}{T_1}=\frac{Q_2}{T_2}. Since T1<T2, Q2>Q1. The difference between Q2 and Q1 is the work done by the heat pump.
 

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