Answer :
In this question, we're dealing with a window air-conditioner unit, and we aim to understand the energy interactions, which is a topic commonly explored in Physics, specifically in thermodynamics.
The air-conditioner unit absorbs heat from the room and transfers it to the outside air. The two quantities involved are:
The amount of heat absorbed from the room is [tex]9.95 \times 10^4 \text{ J/min}[/tex].
The amount of heat deposited into the outside air is [tex]1.54 \times 10^5 \text{ J/min}[/tex].
In a refrigeration or air-conditioning cycle, the unit takes in heat from a cooler space (the room) and releases it to a warmer space (outside). This process requires work input, which is accomplished by the compressor in the air-conditioning system.
To find out how much work the air-conditioner does, we can use the first law of thermodynamics. This law states that the energy added to the system (work done) plus the heat entering the system must equal the heat leaving the system. Mathematically, it's given as:
[tex]Q_{in} + W = Q_{out}[/tex]
Where:
- [tex]Q_{in}[/tex] is the heat absorbed from the room ([tex]9.95 \times 10^4 \text{ J/min}[/tex]).
- [tex]Q_{out}[/tex] is the heat released outside ([tex]1.54 \times 10^5 \text{ J/min}[/tex]).
- [tex]W[/tex] is the work done by the air-conditioner.
Rearranging the equation to solve for [tex]W[/tex], we get:
[tex]W = Q_{out} - Q_{in}[/tex]
Substituting the values:
[tex]W = (1.54 \times 10^5) - (9.95 \times 10^4)[/tex]
[tex]W = 1.54 \times 10^5 - 9.95 \times 10^4 = 5.45 \times 10^4 \text{ J/min}[/tex]
So, the air conditioner does [tex]5.45 \times 10^4 \text{ J/min}[/tex] of work. This is the amount of energy that the air-conditioner requires to transfer the heat from the inside to the outside.
