Answer :
Q(l) = 35,000 kj/h
T° = 4° c = 277k
T(H) = 298 k
given that house is to be maintained at 25° c at all times
hence internal energy is zero. du=0
now, using first law of thermodyanamics
dq= du+dw
dq= dw ( because du = 0)
hence, heat lost by house is equal to the work input for system
Q(l)= w (input)
w (input) =35,000 kj/hr
now, coefficient of performance for reversible heat pump = cop ( rev,hp)
cop ( rev,hp) = = 14.9
cop ( rev,hp) = Q(l)/w (input)
14.9 = 35,000/ (win)rev
(win)rev = 2349 kj/h ( reversible work input)
Irreversibility of process = I
I = (win) - (win)rev
I = 35000- 2349
I = 32651 KJ/hr
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