A heat pump has a coefficient of performance of 4.5. If the heat pump absorbs 46.1 cal of heat from the cold outdoors in each cycle, what is the amount of heat expelled to the warm indoors?
(in cal)
tried everything but got nothing.
CoP=4.5
CoP=Qh/(Qh-Qc)=1/{1-(Qc/Qh)}=4.5
Qc/Qh=0.78
Qc=46.1 cal
Qh= Qc/0.78 = 46.1/0.78=59.1 cal
thanks :)
To find the amount of heat expelled to the warm indoors, you can use the equation:
Qc = Qh / (1 - 1/COP)
Where:
Qc is the amount of heat expelled (in cal)
Qh is the amount of heat absorbed (in cal)
COP is the coefficient of performance
Given that the heat pump absorbs 46.1 cal of heat from the cold outdoors (Qh = 46.1 cal) and the coefficient of performance is 4.5 (COP = 4.5), we can substitute these values into the equation:
Qc = 46.1 cal / (1 - 1/4.5)
Simplifying the equation:
Qc = 46.1 cal / (1 - 2/9)
Qc = 46.1 cal / (7/9)
Qc = 46.1 cal * (9/7)
Qc = 58.71 cal
Therefore, the heat pump expels approximately 58.71 cal of heat to the warm indoors.
To find the amount of heat expelled to the warm indoors by the heat pump, we need to use the formula for the coefficient of performance (COP) of a heat pump.
The coefficient of performance is defined as the ratio of the heat delivered to the warm indoor space to the work input or the heat extracted from the cold outdoor space. Mathematically, it is given by:
COP = Heat delivered / Heat extracted
In this case, the coefficient of performance of the heat pump is given as 4.5, and the heat extracted from the cold outdoors is 46.1 cal. Let's denote the heat expelled to the warm indoors as Q.
We can rearrange the formula to solve for Q:
COP = Q / Heat extracted
Rearranging the formula, we get:
Q = COP x Heat extracted
Substituting the given values, we get:
Q = 4.5 x 46.1 cal
Calculating this value, we find:
Q = 207.45 cal
Therefore, the amount of heat expelled to the warm indoors by the heat pump is 207.45 cal.