Work done by the engine per min
\(\triangle \)W = 5.4 x 10⁸J = 0.54 x 10⁹J/min
Heat absorbed by engine per min
\(\triangle \)Q₁ = 3.6 x 10⁹J/min
\(So,\ engine\ efficiency,\eta =\frac { Work\ done }{ Heat\ delivered }\)
\( \\ =\frac { 0.54\times { 10 }^{ 9 } }{ 3.6\times { 10 }^{ 9 } } =0.15\)
= 15%
Heat wasted per min = heat absorbed - work done
= 3.6 x 10⁹ - 0.54 x 10⁹
= 3.06 x 10⁹ J/min

Given, T₁= -3⁰C = - 3 + 273 = 270K
T₂ = 27⁰C = 27 + 273 = 300K
Efficiency, 
\(\eta =1-\frac { { T }_{ 1 } }{ { T }_{ 2 } } =1-\frac { 270 }{ 300 } =\frac { 1 }{ 10 } \)%
\(or\quad \frac { W }{ Q } =0.5\quad \eta =\frac { 1 }{ 20 }\)
or Q = 20 W = 20kJ per second

Given, T₂ = 9⁰C = 9 + 273 = 282K
T₁ = 36⁰C = 36 + 273 = 309K
Coefficient of performance
\(=\frac { { T }_{ 2 } }{ { T }_{ 1 }-{ T }_{ 2 } } =\frac { 282 }{ 309-282 } =10.4\) 

Carnot cycle is a reversible cyclic process through which heat is converted into mechanical work.

According to second law of thermodynamics, heat energy cannot converted into work completely.

Efficiency of Carnot engine,
η = \(1−\frac{T_2}{T_1}\)
η = \(1−\frac{273 k}{373 k} = \frac{100}{373}\)
= 0.268
= 26.8%

The coefficient of performance of a refrigerator depends on the temperature of source and sink.

All sorts of dissipative forces, e.g force of friction, viscous drag, electrical resistance, non-elasticity, thermal radiation, convection etc.,make a real process irreversible.

Heat rejected by refrigerator remains in the room itself and so, temperature of room increases.Hence, it is false.

As no heat is allowed to be exchanged, the process is adiabatic.
\(\therefore \) \({ \rho }_{ 2 }V_{ 1 }^{ y }\quad ={ \rho }_{ 1 }V_{ 1 }\quad or\quad \frac { { p }_{ 2 } }{ { p }_{ 1 } } =\left( \frac { { V }_{ 1 } }{ { V }_{ 2 } } \right) ^{ y }\)
As \({ V }_{ 2 }=\frac { 1 }{ 2 } { V }_{ 1 }or\frac { { p }_{ 2 } }{ { p }_{ 1 } } =\left( \frac { { V }_{ 1 } }{ { 1 }/{ 2{ V }_{ 1 } } } \right) ^{ 1.4 }={ 2 }^{ 1.4 }=2.64\)

Here, m = 60kg, g = 10m/s², h = 10m
In going up and down once, number of kilocalories burnt
= (mgh + mgh/2) =\(\frac { 3 }{ 2 } \) mgh
 \(=\frac { 3 }{ 2 } \times \frac { 60\times 10\times 10 }{ 4.2\times 1000 } =\frac { 15 }{ 7 } kcal\)
Total number of kilocalories to be burnt for losing 5 kg of weight = 5\(\times \)7000 = 35000 kcal
\(\therefore \) Number of times of the person has to go up and down the stairs 
= \(\frac { 35000 }{ 15/7 } =\frac { 35\times 7 }{ 15 } \times { 10 }^{ 3 }\)  = 16.3 x 10³ times

Leaking of air is adiabatic expansion and adiabatic expansion produces cooling.

If two isothermal curves intersect, this implies that the pressure and volume of a gas are the same at two different temperatures, that's impossible.

When a gas expands, it does work on the surroundings. This work is done on the expense of internal energy and that is why its internal energy and so its temperature decreases.

∴ Coefficient of performance of refrigerator \(=\frac{T_2}{T_1-T_2}=\frac{260}{315-260}=\frac{260}{55}=4.73\)

Here, T₁ = 300, T₂ = 0^o C = 273 K
Heat extracted, Q₂ = mL₁ = 2 kg x 3.33 x 10⁵ J/kg
= 6.66  x 10⁵ J
As, \(\beta =\frac { { Q }_{ 2 } }{ W } =\frac { { T }_{ 2 } }{ { T }_{ 1 }-{ T }_{ 2 } } \)
\(W=\frac { { Q }_{ 2 }({ T }_{ 1 }-{ T }_{ 2 }) }{ { T }_{ 2 } } \)
\(\\ =\frac { 6.66X1{ 0 }^{ 5 }\times(300-273) }{ 273 }\)
\( \\ =65868J\simeq 6.5\times{ 10 }^{ 4 }J\)

A heat engine is a device (or a combination) which converts heat into work.
Its efficiency, \(\eta =\frac { Work\ output }{ Heat\ input } \)
\(\eta =1-\frac { { T }_{ 2 } }{ { T }_{ 1 } } \)
Where, T₂  = temperature of sink
T₁  = temperature of source.
From above expression, we can see that for 100% efficiency, T₂ = 0
It is impossible to design a thermal engine that has 100% efficiency because it is not possible to have a sink with kelvin temperature.

(i)\(\therefore \) process is adiabatic
\(\therefore \)    \(\Delta \)Q = 0
Work done on the gas,\(\Delta \)W = -500J
According to the first law of thermodynamics.
\(\Delta \)Q = \(\Delta \)U +\(\Delta \)W \(\Rightarrow \) \(\Delta \)U = -\(\Delta \)W = 500J
(ii) This is because heat absorbed by a substance (coolant) is directly proportional to the specific heat of the substance.
(iii) This is because in a harbour town, the relative humidity is more than in a desert town. Hence, the climate of a harbour town is without extremes of hot and cold.

Here, heat abs or bed \(=\mathrm{Q}_1=6 \times 10^5 \mathrm{cal}\).
Initial temperature \(=\mathrm{T}_1=227^{\circ} \mathrm{C}=227+273=500 \mathrm{~K}\).
Final temperature \(=\mathrm{T}_2=127^{\circ} \mathrm{C}=127+273=400 \mathrm{~K}\).
As, for Carnot engine;
\( \frac{Q_2}{Q_1}=\frac{T_2}{T_1} \)
\(Q_2=Q_1 \frac{T_2}{T_1} \)
\( \mathrm{Q}_2=\frac{400}{500} \times 6 \times 10^5 \)
\( \mathrm{Q}_2=4.8 \times 10^5 \mathrm{cal} \)
\( \mathrm{Q}_2=\text { Final heat emitted } \)
\( \text { As } \mathrm{w}=\mathrm{Q}_1-\mathrm{Q}_2=6 \times 10^5-4.8 \times 10^5 \)
\(=1.2 \times 10^5 \mathrm{cal} \)
\( \text { Work }=\mathrm{w}=1.2 \times 10^5 \times 4.2 \mathrm{~J} \)
\(\text { Dore }=5.04 \times 10^5 \mathrm{~J} \)

Consider (say \(\mu \) mole) an ideal gas, which is undergoing an adiabatic expansion. Let the gas expands by an infinitesimally small volume dV, at pressure p, then the  infinitesimally small work done given by
dW = pdV
The net work done from an initial volume V₁ is given by
\(W=\int _{ { v }_{ 1 } }^{ { v }_{ 2 } }{ pdV } \)
For an adiabatic process, \(p{ V }^{ \gamma }=constant=K\)
\(\\ p=\frac { K }{ { V }^{ \gamma } } =K{ V }^{ -\gamma }\)
\(\\ \therefore \ W=\int _{ { v }_{ 1 } }^{ { v }_{ 2 } }{ (K{ V }^{ -\gamma })dV } =k\left[ \frac { { V }^{ -\gamma +1 } }{ -\gamma +1 } \right] _{ { v }_{ 1 } }^{ { v }_{ 2 } }\)
\(\\ =\frac { K{ V }_{ 2 }^{ -\gamma +1 }-K{ V }_{ 1 }^{ -\gamma +1 } }{ (1-\gamma ) } \)
\(\\ For\ an\ adiabatic\ process,\)
\( K={ p }_{ 1 }{ V }_{ 1 }^{ \gamma }={ p }_{ 2 }{ V }_{ 2 }^{ \gamma }\)
\(\\ \Rightarrow W=\frac { { { p }_{ 2 }{ V }_{ 2 }^{ \gamma }.{ V }_{ 2 }^{ -\gamma +1 }-{ p }_{ 1 }{ V }_{ 1 }^{ \gamma }.{ V }_{ 2 }^{ -\gamma +1 } } }{ (1-\gamma ) } \)
\(\\ =\frac { 1 }{ (1-\gamma ) } ({ p }_{ 2 }{ V }_{ 2 }-{ p }_{ 1 }{ V }_{ 1 })\)
\(\\ For\ an\ ideal\ gas,\ { p }_{ 1 }{ V }_{ 1 }=\mu R{ T }_{ 1 }\ and\ { p }_{ 2 }{ V }_{ 2 }=\mu R{ T }_{ 2 }.\ So,\ we\ have\)
\(\\ W=\frac { 1 }{ (1-\gamma ) } [\mu R{ T }_{ 2 }-\mu R{ T }_{ 1 }]=\frac { \mu R }{ (1-\gamma ) } [{ T }_{ 1 }-{ T }_{ 2 }]\)