In this page we have Carnot Cycle or engine Problems . Hope you like them and do not forget to like , social share
and comment at the end of the page.
Question 1
A carnot engine has an efficiency of 30%.Its efficiency is to be increased to 50%.By what must the temperature of the source be increased if the sink is at 300 K?
a. 171.4 K
b. 150 K
c. 200 K
d. 160 K
Solution
Efficiency is given by
$\eta = 1 - \frac {T_2}{T_1}$
Here $\eta =30%$ and $T_2=300 K$, $T_1=?$
$.30 = 1 - \frac {300}{T_1}$
or $T_1 =428.6K$
For Increased efficiency of 50%, the source temperature should be
$.30 = 1 - \frac {300}{T_1}$
or $T_1 =600 K$.
Hence the Source Temperature should be increased by 600 -428.6 = 171.4 K
Question 2
In a refrigerator one removes heat from a lower temperature and deposits to the surroundings at a higher temperature. In this process, mechanical work has to be done, which is provided by an electric motor. If the motor is of 1kW power, and heat is transferred from -3°C to 27°C, find the heat taken out of the refrigerator per second assuming its efficiency is 50% of a perfect engine?
a. 18 KJ/s
b. 19 kJ/s
c. 17 KJ/s
d. 16 KJ/s
Solution
Carnot's engine is perfect heat engine operating between two temperature $T_1$ and $T_2$ (source and sink). Refrigerator is also Carnot's engine working in reverse order its efficiency is given by
$ \eta = 1 - \frac {T_2}{T_1} = 1 - \frac {270}{300} = .1$
Efficiency of the refrigerator($\eta _1$)= 50% of .1 = .05
Now Coefficient of performance ($\alpha$} is given by
$\alpha = \frac {1 - \eta _1}{\eta _1} = 19$
Now
$ \alpha = \frac {Q}{W}$
or Q = 19% W.D. by motor on refrigerator
= 19 * 1 kW = 19 kJ/s
Question 3
A system undergoes a cyclic process in which its absorbs $Q_1$ heat and gives out $Q_2$ heat.The efficiency of the process is $\eta$ and the work done is W. Which of the below option are correct
a. $\eta = \frac {W}{Q_1}$
b. $\eta =1 - \frac {Q_2}{Q_1}$
c. $W = Q_1 - Q_2$
d. $\eta = \frac {Q_2}{Q_1}$
Solution
Option (a),(b) and (c) are correct
Question 4
Which of the processes described below are irreversible?
(a) The increase in temperature of an iron rod by hammering it.
(b) A gas in a small container at a temperature $T_1$ is brought in contact with a big reservoir at a higher temperature $T_2$ which increases the temperature of the gas.
(c) A quasi-static isothermal expansion of an ideal gas in cylinder fitted with a frictionless piston.
(d) An ideal gas is enclosed in a piston cylinder arrangement with adiabatic walls. A weight W is added to the piston, resulting in compression of gas
Solution
(a),(b) and (d)
Question 5
A carnot engine operating between Temperature $T_1$ and $T_2$ has the efficiency of $\frac {1}{6}$.When the temperature $T_2$ is lowered by 62K,it efficiency becomes $\frac {1}{3}$,Then Temperature $T_1$ and $T_2$ respectively are
(a) $T_1=370K$,$T_2=330K$
(b) $T_1=330K$,$T_2=268K$.
(c) $T_1=310K$,$T_2=248K$.
(d) $T_1=372K$,$T_2=310K$
Solution
We have,
$ \frac {1}{6} = 1 - \frac {T_2}{T_1}$
and
$ \frac {1}{3} = 1 - \frac {T_2 - 62}{T_1}$
Solving these, we have $T_1=372K$,$T_2=310K$
Question 6
A carnot engine operating between Temperature $T_1=400K$ and $T_2=300K$ .It absorbs 100cal heat from the source Which of the below statement are true?
(a) $\eta = .25$
(b) Work Done by Engine= 104.5 J
(c) Heat Rejected = 70 Cal
(d) $\eta = .30$
Solution
$ \eta = 1 - \frac {T_2}{T_1} = 1 - \frac {300}{400} = .25$
$\text {Work done}= \eta \times \text{heat absorbed} = .25 \times 100 = 25 $Cal = 104.5 J
Question 7
Which of below heat engine can be possible operating between the temperature $T_1= 1400$ and $T_2=300$?
(a) W=1000 J, Q=1083 J
(b) W=800 J, Q=1600 J
(c) W=400 J, Q=705 J
(d) W=600 J, Q=750 J
Solution
The maximum possible Efficiency of an engine is the carnot efficiency
$ \eta = 1 - \frac {T_2}{T_1} = 1 - \frac {300}{1400} = .7857$.
We can calculate Efficiency for each of the possible values. The correct option will be (b) and (c)
Question 8
Consider a heat engine as shown in below figure. $Q_1$ and $Q_2$ are heat added to heat bath $T_1$ and heat taken from $T_2$ in one cycle
of engine. W is the mechanical work done on the engine.
If W > 0, then possibilities are:
(a) $Q_1 > Q_2 > 0$
(b) $Q_2 > Q_1 > 0$
(c) $Q_2 < Q_1 < 0$
(d) $Q_1 < 0$, $Q_2 > 0$
Solution
Option (a, c) are correct
From fig $Q_1 = W + Q_2$
For W > 0 So , $Q_1 - Q_2 > 0$ or $Q_1 > 0$
$Q_1 > Q_2 > 0$ if both $Q_1$, $Q_2$ positive
verifies option (a).
or $Q_2 < Q_1 < 0$ if both $Q_1$, $Q_2$ negative verifies option (c).
Question 9
A freezer has a Coefficient of performance of 5. If the temperature inside the freezer is -20° C. What is the temperature at which its rejects heat.Assume ideal system
(a) 31° C
(b) 41° C
(c) 50° C
(d) 25° C
Solution
$5 = \frac {T_2}{T_1 - T_2} = \frac {253}{T - 253}$
or T=304 K =31 ° C
Question 10
An air conditioner removes heat of 1.5 Kcal/s from a large room. The power required to run it is 2KW. If the typical value of $\frac {Q_2}{W}$ is 1/10 that of a carnot refrigerator and the temperature outside is 313 K. What is the temperature inside the room?
(a) 31° C
(b) 30° C
(c) 29° C
(d) 25° C
Solution
Carnot refrigerator is given as
$(\frac {Q_2}{W})_{carnot} = \frac {T_2}{T_1-T_2}$
Now for air conditioner
$\frac {Q_2}{W} = \frac {1}{10} (\frac {Q_2}{W})_{carnot} =\frac {1}{10} \frac {T_2}{T_1-T_2}$
By rearranging
$T_2 = \frac {10Q_2 T_1}{W+ 10 Q_2}$ --(1)
Here $T_1=273 + 40 =313 K$
Power=2 KW,So
W=2000 J = .478 Kcal
$Q_2=1.5 Kcal$
Substituting all the values
$T_2= 303 K$=30° C
link to this page by copying the following text
Also Read
- Notes
- Assignments
- Revision Notes
Go back to Class 11 Main Page using below links
Class 11 Maths
Class 11 Physics
Class 11 Chemistry
Class 11 Biology