Two carnot air conditioners, a and b, are removing heat from different rooms. the outside temperature is the same for both rooms, 313 k. the room serviced by unit a is kept at a temperature of 294 k, while the room serviced by unit b is kept at 302 k. the heat removed from either room is 4510 j. for both units, find (a), (b) the magnitude of the work required and (c), (d) the magnitude of the heat deposited outside for a and b conditioners respectively.

Explanation:

Coefficient of performance of cooling apparatus

= Heat output / energy input

= T₁ / T₁ - T₂

Where T₁ = Temp of hot end and T₂ = cold end

For A

Heat output / energy input = 313 / (313 - 294)

= (Heat withdrawn + energy input )/ energy input = 313 / 19

1 + Heat withdrawn/ energy input = 313/ 19

Heat withdrawn/ energy input = 294 / 19

energy input = 19/294 x Heat withdrawn

= 19/294 x 4510

= 291.46 J

Heat deposited = 291.46 + 4510 = 4801.46 J

Similarly for B

energy input = 313-302 /302 x Heat withdrawn

= 11 / 302 X 4510

= 164.27

Heat deposited = 164.27 + 4510 = 4674.27 J

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west