Air conditioners sold in the united states are given a seasonal energy-efficiency ratio (seer) rating that consumers can use to compare different models. a seer rating is the ratio of heat pumped to energy input, similar to a cop but using english units, so a higher seer rating means a more efficient model. you can determine the cop of an air conditioner by dividing the seer rating by 3.4for inside temperature 25? c and outside temperature 34? c when you'd be using air conditioning, estimate the theoretical maximum seerrating of an air conditioner. (new air conditioners must have a seer rating that exceeds 13, quite a bit less than the theoretical maximum, but there are practical issues that reduce efficiency.)

112.58

Explanation:

The Coefficient of Performance of any system is denoted by COP=Q/W, where Q is the useful heat supplied or removed and W is the work required by the system. According to the first law of thermoddynamics Qh= Qc + W, where Qh is the heat transfered to the hot reservoir and Qc is the heat collected from the cold reservoir. Substituting the values for W and apllying the limitation for maximum theoretical efficiency we end up with the eqution shown below.

The Coefficient of Performance of air conditioner or COP is denoted by

COP(cool) = Tc/(Th- Tc)

where Tc: the lowest temperature

           Th: the highest temperature

converting the values to Kelvin and adding them in the above equation

COP(cool) = (25+273)/((34+273)-(25+273))

                 = 298/(307-298)

                 = 298/9 = 33.11

From the question, it is stated that COP=SEER/3.4

hence, SEER= COP * 3.4

SEER= 33.11 * 3.4 = 112.58

the independent variable, or the one that is intentionally manipulated is the speed.

ok

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