Physics · Heat and Thermodynamics
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A reversible Carnot engine converts 1/6th of heat into input work. When the temperature of the sink is reduced by 62˚C and the efficiency is doubled then the temperature of the source and sink is:
- A
80oC ,37oC
- B
99oC , 30oC
- C
99oC,25oC
- D
99oC , 37oC
Efficiency of Engine (),
= Work / Qinput
As, W = 1 / 6 Qinput
= 1 / 6
Also, = 1 - T2 / T1
Where
T1 and T2 are the temperatures of the source and sink respectively.
So,
1 / 6 =1 - T2 / T1
T1 = 1.2 T2 ………… (1)
Now the sink temperature is reduced by 62C, i.e. T2’ = T2 - 62 ………… (2)
Also the new efficiency is doubled, i.e. ’ = 2 1 / 6 = 2 / 6 = 1 / 3 …… (3)
Substituting (1), (2) and (3)
’ = 1 - T2’ / T1
1 / 3 = 1 - (T2 - 62) / 1.2 T2
T2 = 310 K
Thus T2 = 310 - 273 = 37C
From (1), T1 = 1.2 T2
T1 = 1.2 x 310 = 372 K
i.e T1 = 372 - 273 = 99C
Efficiency of Engine (),
= Work / Qinput
As, W = 1 / 6 Qinput
= 1 / 6
Also, = 1 - T2 / T1
Where
T1 and T2 are the temperatures of the source and sink respectively.
So,
1 / 6 =1 - T2 / T1
T1 = 1.2 T2 ………… (1)
Now the sink temperature is reduced by 62C, i.e. T2’ = T2 - 62 ………… (2)
Also the new efficiency is doubled, i.e. ’ = 2 1 / 6 = 2 / 6 = 1 / 3 …… (3)
Substituting (1), (2) and (3)
’ = 1 - T2’ / T1
1 / 3 = 1 - (T2 - 62) / 1.2 T2
T2 = 310 K
Thus T2 = 310 - 273 = 37C
From (1), T1 = 1.2 T2
T1 = 1.2 x 310 = 372 K
i.e T1 = 372 - 273 = 99C
Efficiency of Engine (),
= Work / Qinput
As, W = 1 / 6 Qinput
= 1 / 6
Also, = 1 - T2 / T1
Where
T1 and T2 are the temperatures of the source and sink respectively.
So,
1 / 6 =1 - T2 / T1
T1 = 1.2 T2 ………… (1)
Now the sink temperature is reduced by 62C, i.e. T2’ = T2 - 62 ………… (2)
Also the new efficiency is doubled, i.e. ’ = 2 1 / 6 = 2 / 6 = 1 / 3 …… (3)
Substituting (1), (2) and (3)
’ = 1 - T2’ / T1
1 / 3 = 1 - (T2 - 62) / 1.2 T2
T2 = 310 K
Thus T2 = 310 - 273 = 37C
From (1), T1 = 1.2 T2
T1 = 1.2 x 310 = 372 K
i.e T1 = 372 - 273 = 99C
Efficiency of Engine (),
= Work / Qinput
As, W = 1 / 6 Qinput
= 1 / 6
Also, = 1 - T2 / T1
Where
T1 and T2 are the temperatures of the source and sink respectively.
So,
1 / 6 =1 - T2 / T1
T1 = 1.2 T2 ………… (1)
Now the sink temperature is reduced by 62C, i.e. T2’ = T2 - 62 ………… (2)
Also the new efficiency is doubled, i.e. ’ = 2 1 / 6 = 2 / 6 = 1 / 3 …… (3)
Substituting (1), (2) and (3)
’ = 1 - T2’ / T1
1 / 3 = 1 - (T2 - 62) / 1.2 T2
T2 = 310 K
Thus T2 = 310 - 273 = 37C
From (1), T1 = 1.2 T2
T1 = 1.2 x 310 = 372 K
i.e T1 = 372 - 273 = 99C
Efficiency of Engine (),
= Work / Qinput
As, W = 1 / 6 Qinput
= 1 / 6
Also, = 1 - T2 / T1
Where
T1 and T2 are the temperatures of the source and sink respectively.
So,
1 / 6 =1 - T2 / T1
T1 = 1.2 T2 ………… (1)
Now the sink temperature is reduced by 62C, i.e. T2’ = T2 - 62 ………… (2)
Also the new efficiency is doubled, i.e. ’ = 2 1 / 6 = 2 / 6 = 1 / 3 …… (3)
Substituting (1), (2) and (3)
’ = 1 - T2’ / T1
1 / 3 = 1 - (T2 - 62) / 1.2 T2
T2 = 310 K
Thus T2 = 310 - 273 = 37C
From (1), T1 = 1.2 T2
T1 = 1.2 x 310 = 372 K
i.e T1 = 372 - 273 = 99C
Tagged under Physics · Heat and Thermodynamics · 2021