The energy extracted from the cold reservoir is 970 Joule. The result is obtained by using the formula of coefficient of performance for a heat pump.
The coefficient of performance is a ratio between energy produced and energy used. It expressed the efficiency of a heat pump. COP of a heat pump can be expressed as
[tex]COP = \frac{Q_{h} }{W} = \frac{Q_{h} }{Q_{h} - Q_{c}}[/tex]
Where
If
Then, what is the value of Qc?
The value of Qc is
[tex]COP = \frac{Q_{h} }{Q_{h} - Q_{c}}[/tex]
[tex]12.7 = \frac{1,053 }{1,053 - Q_{c}}[/tex]
[tex]12.7 (1,053 - Q_{c}) = 1,053[/tex]
[tex]13,373.1 - 12.7Qc = 1,053[/tex]
[tex]13,373.1 - 1,053 = 12.7Qc[/tex]
[tex]12,320.1 = 12.7Qc[/tex]
[tex]Qc = \frac{12,320.1}{12.7}[/tex]
[tex]Qc = 970 Joule[/tex]
Hence, the energy extracted from the cold reservoir of the heat pump is 970 Joule.
Learn more about COP of heat pump here:
https://brainly.com/question/13185041
#SPJ4