Question

two carnot engines A and B are operated in series.engine A receives heat from a reservoir at 600K and rejects heat to a reservoir at temperature T.engine B receives heat rejected by engine A and inturn rejects heat to a reservoir at 100K. if the efficiencies of two engines A and B are represented by nA and nB respectively,then what is the value of nB/nA

Answer 1

it jee mains mock paper quez
we know tat fr a reserviour
neta=heat absorbed /work done
ie, netaA=t1/(t1+1) where t1 is temp
for 1st case when 600k is supplied
similarly fr netaB=t2/(t2+1)
so ratio will be 12/7

Answer 2

Hey buddy,

◆ Answer-
nB / nA = 12/7

◆ Explanation-
# Given-
T1 = 600 K
T2 = ?
T3 = 100 K

# Solution-
As carnot engines are connected in series work done will be same -
W1 = Q2 - Q3
W2 = Q1 - Q2

Therefore,
W1 = W2
Q1 - Q2 = Q2 - Q3
Q2 = (Q1 + Q3) / 2

In carnot cycle, Q ∝ T, hence
T2 = (T1 + T3) / 2
T2 = (600 + 100) / 2
T2 = 350.

Efficiency of 1st engine -
nA = (Q1 - Q2) / Q1
nA = (T1 - T2) / T1
nA = (600 - 350) / 600
nA = 5/12

Efficiency of 1st engine -
nB = (Q2 - Q3) / Q2
nB = (T2 - T3) / T2
nB = (350 - 100) / 350
nB = 5/7

Value of nB/nA -
nB / nA = (5/7) / (5/12)
nB / nA = 12/7

Hope this helps you...