Second law of thermodynamics | Numerical problems and short answer questions collection | Grade 12 NEB (Physics) | Physics in Depth
Second law of thermodynamics
Kelvin's statement: It is impossible to get a continuous supply of work from a body by cooling it to a temperature lower than that of its surroundings.
Clausius statement: It is impossible for a self acting machine unaided by any external agency to convey heat from one body to another at a higher temperature.
Kelvin's statement can also be stated more precisely as : It is not possible to design a heat engine which works in a cyclic process and whose only result is to take heat from a body at a single temperature and convert it completely into mechanical work.
Heat and work are not equivalent! . One can convert mechanical work completely into heat but one cannot convert heat completely into mechanical work.
Short Answer Questions Solution
- State Second law of thermodynamics. Explain its significance.
Solution :
Second law of thermodynamics can be stated as :
Kelvin's statement: It is impossible to get a continuous supply of work from a body by cooling it to a temperature lower than that of its surroundings.
Clausius statement: It is impossible for a self acting machine unaided by any external agency to convey heat from one body to another at a higher temperature.
Significance of second law of thermodynamics:
- It indicates the direction on heat transfer i.e., it clearly states that heat flows from higher temperature to lower temperature when unaided by external agaency.
- It indicates the non - equivalence of heat and work i.e., one can convert mechanical work completely into heat but one cannot convert heat completely into mechanical work.
- It states that the entropy of an isolated system cannot decrease.
- Will the temperature of room decrease if we leave the door of refrigerator open?
Solution :
No, the temperature of room will not decrease. Refrigerator is a heat engine that works in reverse manner i.e., it transfers heat from a cold body to hot body. Since, the room is at higher temperature, the refrigerator rejects heat to the room and warms the room. - Is entropy equal to zero in adiabatic process? Explain.
Solution :
Change of entropy of a system is defined as, \[dS=\frac{dQ}{T}\] Since, during the adiabatic process, there is no heat transfer between the system and the surrounding (i.e., dQ = 0). So, dS = 0. Though the entropy is not equal to zero in adiabatic process, there is no additional increase in the process i.e., entropy remains constant. - Why do diesel engines need no spark plugs?
Solution :
In a diesel engine, the air is compressed adiabatically until its temperature rises about 700 \(^\circ\) C. When a spray of diesel oil is injected into the cylinder of the engine, the diesel oil burns producing large amount of energy. Thus, no spark plugs is needed for diesel engine. - Explain the significant differences between first law and second law of thermodynamics.
Solution :
The differences between the first law and second law of thermodynamics are :
First law of thermodynamics :
- It does not indicate the direction of heat transfer.
- It does not tell to what extent the mechanical energy can be obtained from the heat energy.
- It clearly states that the direction of heat transfer takes place from higher temperature to lower temperature unless an external work is done on the system.
- It helps to calculate the efficiency of the heat engine i.e., what extent of mechanical energy can be obtained from the heat energy.
- Can the efficiency of petrol/diesel engine be equal to 100 % ? Explain.
Solution :
The efficiency of heat engine like petrol / diesel is given by \[\eta=\frac{Q_1-Q_2}{Q_1}\] Here, \(Q_1-Q_2\) is the heat converted to the mechanical work. For engines to have 100 % , the output work should be equal to the input heat energy given to the system. However, some of the heat is always lost as a friction or by other means, so efficiency of petrol/diesel engine can never be equal to 100 %. - Petrol engine is less efficient than the diesel engine. Why?
Solution :
The efficiency of petrol an diesel engine is given by \[\eta=1-\left(\frac{1}{\rho}\right)^{\gamma-1}\] Where \(\rho\) is the compression ratio and \(\gamma\) is the ratio of two specific heats of a gas.
Since the compression ratio of diesel engine is greater than the petrol engine, diesel engine is more efficient than petrol engine. - Can the carnot engine be generalized in practice?
Solution :
No. Carnot engine is an ideal engine whose conditions cannot be realized in practice (i.e., it has ignored the friction between the cylinder and the piston, gas is supposed to be ideal, operations on the gas are supposed to be performed very quickly, etc. ).
Numerical Problems Solution
- In a petrol engine the rate of production of heat due to combustion of fuel is 7.46\(\times\)10\(^5\) cal/hr. If the efficiency of the engine is 30\(\%\), calculate the horse power of the engine.
Solution :
In 1 hour, the heat supplied to the engine is 7.46\(\times\)10\(^5\) cal i.e., 4.2 \(\times\) 7.46 \(\times\) 10\(^5\) J
Since, efficiency of engine is 30\(\%\), work output of the engine in 1 hour is \[\begin{align*} &=30 \% \hspace{0.1cm}\text{of}\hspace{0.1cm} 4.2 \times 7.46 \times 10^5\\ &=939960 \hspace{0.1cm}\text{J} \end{align*}\] Now, the power of engine is \[\begin{align*} \text{Power}&=\frac{\text{Work output}}{\text{time}}\\ &=\frac{939960}{3600}\\ &=261.1 \hspace{0.1cm} \text{W}\\ &=\frac{261.1}{740}\hspace{0.1cm} \text{H.P.}\\ &=0.35\hspace{0.1cm} \text{H.P.} \end{align*}\] - A carnot engine whose low temperature reservoir is at 280 K has an efficiency of 40 % . It is desired to increase this to 50 %. By how much many degrees must the temperature of the low temperature reservoir be decreased
if that of the high temperature reservoir remains constant.
Solution :
lower temperature, \(T_2\) = 280 K
initial efficiency of machine, \(\eta_1\) = 40 \(\%\)
final efficiency of machine, \(\eta_2\) = 50 \(\%\)
higher temperature, \(T_1\) = x (say)
final lower temperature, \(T_2'\) = ?
decrease in lower temperature, \(T_2-T_2'\) = ?
Now, \[\begin{align*} \eta_1&=\left(1-\frac{T_2}{T_1}\right)\times 100 \%\\ 40&=\left(1-\frac{280}{x}\right)\times 100\\ \frac{2}{5}&=1-\frac{280}{x}\\ \frac{280}{x}&=\frac{3}{5}\\ x&=\frac{280\times 5}{3}\\ \therefore x&=466.67 \hspace{0.1cm}\text{K} \end{align*}\] Then, \[\begin{align*} \eta_2&=\left(1-\frac{T_2'}{T_1}\right)\times 100 \%\\ 50&=\left(1-\frac{T_2'}{466.67}\right)\times 100\\ 0.5&=\left(1-\frac{T_2'}{466.67}\right)\\ \frac{T_2'}{466.67}&=0.5\\ \therefore T_2'&=233.33 \hspace{0.1cm}\text{K} \end{align*}\] Thus, the lower temperature should be decreased by \(T_2-T_2'\) = 280 - 233.33 K = 46.67 K. - A diesel engine performs 2200 J of mechanical work and discards 4300 J of heat each cycle. (i) How much heat must be supplied to the engine in each cycle? (ii) What is the thermal efficiency of the engine?
Solution :
work done by the engine, W = 2200 J
heat rejected by the engine, \(Q_2\) = 4300 J
heat supplied to the engine, \(Q_1\) = ?
thermal efficiency of the engine, \(\eta\) = ?
For heat engine, \[W=Q_1-Q_2\] Then,
(a)
\[\begin{align*} W&=Q_1-Q_2\\ 2200&=Q_1-4300\\ Q_1&=6500 \hspace{0.1cm}\text{J} \end{align*}\] (b)
\[\begin{align*} \eta&=\frac{Q_1-Q_2}{Q_1}\times 100\%\\ &=\frac{2200}{6500}\times 100\%\\ \therefore \eta&= 33.85 \% \end{align*}\] - A Carnot engine works between 800 \(^\circ\) C and 400 \(^\circ\) C. If it is possible either to increase the source temperature by 50\(^\circ\) C or to decrease the sink temperature by 50 \(^\circ\) C, which of these actions will be causing more increase in the efficiency? Justify your answer.
Solution :
Initial case,
sink temperature, \(T_2\) = 400 \(^\circ\) C = 673 K
source temperature, \(T_1\) = 800 \(^\circ\) C = 1073 K
When the source temperature increases by 50 \(^\circ\) C , then
Source temperature, \(T_1'\) = 850 \(^\circ\) C = 1123 K
So, the efficiency is,
\[\begin{align*} \eta&=\left(1-\frac{T_2}{T_1'}\right)\times 100\%\\ &=\left(1-\frac{673}{1123}\right)\times 100\%\\ &=40.07\% \end{align*}\] When the sink temperature decreases by 50 \(^\circ\) C , then
sink temperature, \(T_2'\) = 350 \(^\circ\) C = 623 K
So, the efficiency is
\[\begin{align*} \eta'&=\left(1-\frac{T_2'}{T_1}\right)\times 100\%\\ &=\left(1-\frac{623}{1073}\right) \times 100 \%\\ &=41.93 \% \end{align*}\] It is more efficient to decrease the sink temperature by 50 \(\circ\) C because the efficiency in this case is higher (i.e., 41.93 %) - A petrol engine consumes 10 kg of petrol in one hour. The calorific value of petrol is 11.4 \(\times\) 10\(^3\) cal/gm. The power of the engine is 20 KW. Calculate the efficiency of the engine.
Solution :
The calorific value of petrol is 11.4 \(\times\) 10\(^3\) cal/gm.
1 gm of petrol produces 11.4 \(\times\) 10\(^3\) cal energy.
1000 gm (i.e., 1 kg) of petrol produces 11.4 \(\times\) 10 \(^3\) \(\times\) 4.2 \(\times\) 1000 J energy . \((\because 1 \hspace{0.1cm}\text{cal}=4.2\hspace{0.1cm} J)\)
10 kg of petrol produces 11.4\(\times\) 10 \(^3\) \(\times\) 4.2 \(\times\) 10 \(\times\) 1000 J i.e., 478800000 J
So, the input power is
\[\begin{align*} P_1&=\frac{478800000}{3600}\hspace{0.1cm} \because \text{Power}=\frac{\text{Work}}{\text{time}}\\ P_1&=133000 \hspace{0.1cm}\text{W} \end{align*} \] Here, output power is \[P_2 = 20000 \hspace{0.1cm}\text{W}\] So, the efficiency of engine is
\[\begin{align*} \eta&=\frac{P_2}{P_1}\times 100\%\\ &=\frac{20000}{133000}\times 100\%\\ \therefore \eta&=15.04\% \end{align*}\]
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