Answer:
14.07 units of joules is removed for each joules of electricity used
Explanation:
The heated room temperature [tex]T_{c}[/tex] = 61 °F
The outside temperature [tex]T_{h}[/tex] = 98 °F
For conversion from fahrenheit to kelvin we use the equation
(32°F − 32) × 5/9 + 273.15
[tex]T_{c}[/tex] = 289.26
[tex]T_{h}[/tex] = 309.82
For air conditioning,
COP = [tex]\frac{T_{c} }{T_{h} -T_{c} }[/tex]
COP = [tex]\frac{289.26 }{309.82 -289.26 }[/tex] = 14.07
This means that 14.07 units of joules is removed for each joules of electricity used