Cancel R/M:
Q_dot = V_dot_1*(P1/T1) * ( k/(k-1) ) * (T2 - T1)
SAVE THIS FORMULA for later.
Part B:
This part wants us to assume a Carnot cycle (the most perfect performing cycle that is possible to exist) acts as the refrigeration of the great outdoors, and the consequential heating of the house via the heat rejection component.
We have a formula for Carnot coef of performance of a refrigerator.
COPr_carnot = 1/(T_H/T_L - 1)
Important: T_H and T_L absolutely MUST be in the Kelvin or Rankine scale.
However, this isn't a refrigerator per-say, because we aren't desiring to cool off the great outdoors. We are desiring to heat the house. This changes the definition of "what we want", and thus will change the formula that will yield the ratio of "what we want"/"what we pay for".
This is still the same gismo as a refrigerator, we've just switched the inside and outside. Still, use the same formula for energy balance on it:
Q_in + W_in = Q_out
Refrigeration "cooling" COP:
COPr = Q_in/W_in
"what we want" = Q_in, heat "sucked" from the interior of the house in to the device.
Heat pump "warming" COP:
COPh = Q_out/W_in
"what we want" = Q_out, heat rejected from the device in to the house
COPh = (Q_out)/W_in
COPh = (Q_in + W_in)/W_in
COPh = COPr + 1
COPh_carnot = 1/(T_H/T_L - 1) + 1
COPh_carnot = 1/(1 - T_L/T_H)
From earlier, we know the answer to "what we want", which we've called Q_dot then. Call it Q_dot_out, now, because it is heat expelled by the heat pump device. Compare it with what we pay for, W_dot_in
COPh = Q_dot_out/W_dot_in
Solve for W_dot_in:
W_dot_in = Q_dot_out/COPh
Best case scenario, COPh = COPh_carnot
Thus:
W_dot_in = Q_dot_out * (1 - T_L/T_H)
Summary:
A: Q_dot = V_dot_1*(P1/T1) * ( k/(k-1) ) * (T2 - T1)