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Practice each question, then open the answer or worked solution.
Consider a heat engine as shown on the figure. Q1 and Q2 are heat added to heat bath T1 and heat taken from T2 in one cycle of the engine. W is the mechanical work done on the engine.
If W > 0, then possibilities are:
a) Q1 > Q2 > 0
b) Q2 > Q1 > 0
c) Q2 < Q1 < 0
d) Q1 < 0, Q2 > 0
The correct answers are
a) Q1 > Q2 > 0
c) Q2 < Q1 < 0
Very Short AnswersCan a system be heated and its temperature remains constant?
For temperature to remain constant, the work done by the system against the surrounding should compensate with the heat that is supplied.
According to the given information,
Therefore,
A system goes from P to Q by two different paths in the P-V diagram as shown in the figure. Heat given to the system in the path 1 is 1000 J. The work done by the system along path 1 is more than path 2 by 100 J. What is the heat exchanged by the system in path 2?
From path 1, Q1 = +1000 J
Work done = W1 – W2 = 100
Where,
W1 is the work done through path 1
W2 is the work done through path 2
Therefore, W2 = W1 – 100
Substituting the values, we get Q2 = 900 J
If a refrigerator’s door is kept open, will the room becomes cool or hot? Explain.
If a refrigerator’s door is kept open, the room becomes hot because the amount of heat absorbed by the refrigerator and the work done by the refrigerator will be rejected to the room.
Is it possible to increase the temperature of a gas without adding heat to it? Explain.
Yes, it is possible to increase the temperature of a gas without adding heat to it. This is possible during adiabatic compression.
Air pressure in a car tyre increases during driving. Explain.
Air pressure in a car tyre increases during driving while the volume remains constant which is based on the Charle’s law. Pressure is proportional to the temperature. Therefore, the pressure of the gas increases.
Short AnswersConsider a Carnot’s cycle operating between T1 = 500K and T2 = 300K producing 1 kJ of mechanical work per cycle. Find the heat transferred to the engine by the reservoirs.
Given,
Temperature of the source, T1 = 500 K
Temperature of the sink, T2 = 300 K
Work done per cycle, W = 1 kJ = 1000 J
Heat transferred to the engine per cycle, Q1 =?
Efficiency of a Carnot engine
= 2500 J – 1000 J = 1500 J
A person of mass 60 kg wants to lose 5 kg by going up and down a 10 m high stairs. Assume he burns twice as much fat while going up than coming down. If 1 kg of fat is burnt on expending 7000 kilocalories, how many times must he go up and down to reduce his weight by 5 kg?
Given,
Height of the stairs, h = 10 m
Work done to burn 5 kg of fat = (5)(7000 × 103)(4.2) = 147 × 106 J
Work done towards burning of fat in one trip = mgh + 1/2 mgh = 3/2 mgh = (3/2)(60)(10)(10) = 9 × 103 J
Therefore, no.of trips required = 16.3 × 103 times.
Consider a cycle tyre being filled with air by a pump. Let V be the volume of the tyre and at each stroke of the pump ∆V of air is transferred to the tube adiabatically. What is the work done when the pressure in the tube is increased from P1 to P2?
Following is the equation before and after the stroke:
Therefore, work done is given as
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 -3oC to 27oC, find the heat taken out of the refrigerator per second assuming its efficiency is 50% of a perfect engine.
Efficiency of the Carnot engine is
Where
T1 = 300 K
T2 = 270 K
It is given that efficiency is 50% = 0.5
Efficiency of refrigerator = 0.05
We know that
Therefore, the efficiency of the refrigerator is
