(i) Anoop possesses the qualities like having scientific attitude, awareness, intelligence and keen observer.
(ii) The relative humidity in a harbour town is more than that in a town in a desert. Hence the climate of a harbour town is more temperate than that of a town in a desert.

Here, steam point,
T₁ = 100 ⁰C = 100 + 273 = 373 K
ice point, T₂ = 0°C = 0 + 273 = 273 K
As \(η=1-{T_2\over T_1}\)
\(∴\ \ η=1-{273\over 373}={100\over 373}\)
\(={100\over 373}\times100\%=26.81\%\)

To increase the efficiency [= 1 - (1/ρ)^⋎-1], the compression ratio p has to be increased. ρ cannot be made greater than 10, because then the cylinder of the engine will have to be made very thick and heavy which will be unsuitable for lighter vehicles. Secondly, during the adiabatic compression, the temperature of the air-petrol mixture will be high enough to cause explosion.

Here P₁ = 4 atm, P₂ = 1 atrn, T₁ - 27°C = 300 K and ⋎ = 1.4
The sudden burst of tyre is an adiabatic process, in which
\(P_1^1-\ ^γT^γ_2 = \ ^γT^γ_2\)
\(T_2=T_1\left( P_1\over P_2\right)^{1-⋎\over ⋎}= T_1\left(P_2\over P_1\right)^{⋎-1\over ⋎}\)
\(=300\left(4\over 1\right)^{1.4-1\over 1.4}=201.9\)
= 202K or -71⁰C.

Under isothermal conditions, \(K_i={ΔP\over ΔV_1/V}=P\)
under adiabatic condition, \(K_a={ΔP\over ΔV_2/V}=γP\)
Dividing (ii) by (i), we get
\({ΔV_1\over ΔV_2}=⋎.\ As\ ⋎>1\)
(ΔV₁) > (ΔV₂)

Here heat required to raise temperature of 1 mole of gas through 10 K under constant pressure conditions ΔQ 207 J
\(∴\ \ C_p={ΔQ\over μ.ΔT}={207\over 1\times10}=20.7J\ mol^{-1}K^{-1}\)
∴ C_v = C_p - R = 20.7 - 8.3 = 12.4 J mol^-1 K^-1
∴ Amount of heat required to raise the temperature of gas through 10 K under constant volume condition:
ΔQ' = μ.C_p.ΔT = 1 x 12.4 x 10 = 124 J mol^-1 K^-1

(i) Efficiency of A = efficiency of B
\({ \eta }_{ A }={ \eta }_{ B }\)
\(\\ \Rightarrow 1-\frac { T }{ 900 } =1-\frac { 400 }{ T } \)
\(\\ \Rightarrow { T }^{ 2 }=900\times400\)
\(\\ T=600K\)
(i) Let the first engine take Q₁ heat as input at temperature,
T₁ = 800 K
and gives out heat Q₂ at temperature T₀ The second engine receive Q₂ as input and give is out heat Q₃ at temperature T₃ = 300 K to the sink.
Work done by first (A) engine = work done by second (B) engine.
Thus, Q₁ - Q₂ = Q₂ - Q₃
Dividing both sides by Q₁
\(1-\frac { { Q }_{ 2 } }{ { Q }_{ 1 } } =\frac { { Q }_{ 2 } }{ { Q }_{ 1 } } -\frac { { Q }_{ 3 } }{ { Q }_{ 1 } } \)
\(\\ \Rightarrow 1-T/{ T }_{ 1 }=\frac { { Q }_{ 2 } }{ { Q }_{ 1 } } (1-\frac { { Q }_{ 3 } }{ { Q }_{ 2 } } )\)
\(\\ \Rightarrow -T/{ T }_{ 1 }=\frac { T }{ { T }_{ 1 } } (1-{ T }_{ 3 }/T)\)
\(\\ \Rightarrow { T }_{ 1 }/T-1=1-\frac { { T }_{ 3 } }{ T }\)
\( \Rightarrow \frac { { T }_{ 1 } }{ T } +\frac { { T }_{ 3 } }{ T } =2\)
\(\\ \Rightarrow \frac { 1 }{ T } ({ T }_{ 1 }+{ T }_{ 3 })=2\)
\(\Rightarrow T=\frac { { T }_{ 1 }+{ T }_{ 3 } }{ 2 } \)
\(\\ \Rightarrow T=\frac { 900+400 }{ 2 } =650K\) 

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) Q₂ = 5J
(ii) Coefficient of performance
\(\beta =\frac { { T }_{ 2 } }{ { { T }_{ 1 }-{ T }_{ 2 } } } =\frac { 250 }{ 300-250 } =5\)

Given, work done (W) = - 22.3 J
Work done is taken negative as work is done on the system.
In an adiabatic change, \(\Delta \)Q = 0
Using first law of thermodynamics
\(\Delta \).U = \(\Delta \)Q - W = 0 -(- 22.3) = 22.3J
For another process between states A and B,
Heat absorbed (\(\Delta \)Q) = + 9.35 cal
= + (9.35 x 4.19) J = + 39.18 J
Change in internal energy between two states via different paths are equal.
\(\because \)   \(\Delta \)U = 22.3 J
\(\therefore \)   From first law of thermodynamics,
\(\Delta \)U = \(\Delta \)Q - W
or W = \(\Delta \)Q -\(\Delta \)U
= 39.18 - 22.3
= 16.88J \(\approx \)16.9J

(a) When two bodies at different temperatures T₁ and T₂ are brought in thermal contact, heat flows from the body at the higher temperature to the body at the lower temperature till equilibrium is achieved, i.e., the temperatures of both the bodies become equal. The equilibrium temperature is equal to the mean temperature (T₁ + T₂)/2 only when the thermal capacities of both the bodies are equal.
(b) The coolant in a chemical or nuclear plant should have a high specific heat. This is because higher the specific heat of the coolant, higher is its heat-absorbing capacity and vice versa. Hence, a liquid having a high specific heat is the best coolant to be used in a nuclear or chemical plant. This would prevent different parts of the plant from getting too hot.
(c) When a car is in motion, the air temperature inside the car increases because of the motion of the air molecules. According to Charles’ law, the temperature is directly proportional to pressure. Hence, if the temperature inside a tyre increases, then the air pressure in it will also increase.
(d) A harbour town has a more temperate climate (i.e., without the extremes of heat or cold) than a town located in a desert at the same latitude. This is because the relative humidity in a harbour town is more than it is in a desert town.

887⁰C