Yes, for example in adiabatic compression

Yes, internal energy of lead increases. No heat energy from outside enters the lead

The mechanical work can be completely converted into heat, but heat extracted from some body cannot be completely converted into useful work.

Because work done in compressing the gas increases the internal energy of the gas.

The efficiency is independent of the nature of the working substance.

Change in pressure will be more than 3 times the initial pressure.

Yes, for example in an adiabatic expansion and explosion of a bomb (chemical energy is converted into K.E.).

No, the coefficient of performance of refrigerator decreases with decrease in its inside temperature.

Zero, because for an ideal gas internal energy is wholly kinetic and it is a function of temperature. As temperature remains constant in an isothermal process, hence, internal energy of an ideal gas remains constant.

Given, T₁ = 600K, T₂ = 200K
Efficiency of Carnot engine,
\(\eta =\frac { { T }_{ 1 }-{ T }_{ 2 } }{ { T }_{ 1 } } =\frac { 600-200 }{ 600 } \)
\(=\frac { 400 }{ 600 } =\frac { 2 }{ 3 } =66\)
\(Real\ efficiency=\frac { Energy\ output }{ Energy\ input } =\frac { 2 }{ 1\times 4.2 } =0.47=47%\)
\(\\ \therefore \frac { Real\ efficiency }{ Carnot\ engine\ efficiency } =\frac { 47 }{ 66 } =0.71\)
\(\therefore \frac { Real\ efficiency }{ Carnot\ engine\ efficiency } =\frac { 47 }{ 66 } =0.71\)

According to the second law of thermodynamics whole of the heat cannot be converted completely into work. Some part of the heat that is not converted into work is released by the engine to the atmosphere /9as sink).
Thus, thermal pollution will always occur.

No, it is not constant quantity, as inside, temperature of the refrigerator decreases, it is coefficient of performance also decreases.

Because in the hilly area, temp of surrounding is lower than that of plains.
As \(\eta =1-\frac { { T }_{ 2 } }{ { T }_{ 1 } } \)

Here, mass of gas, m =  2\(\times \)10^-2 kg = 20g
Rise in temperature, \(\Delta \)T = 45^oC
Heat required,\(\Delta \)Q = ?
Molecular mass, M =28
Number of moles, n =\(\frac { m }{ n } =\frac { 20 }{ 28 } =0.714\)
As nitrogen is a diatomic gas, molar specific heat at constant pressure is
\(C_{ p }=\frac { 7 }{ 2 } R=\frac { 7 }{ 2 } \times 8.3J\quad mol^{ -1 }K^{ -1 }\)
\(\\ As\ \Delta Q=nC_{ p }\Delta T\)
\(\\ \therefore \ \Delta Q=0.714\times \frac { 7 }{ 2 } \times 8.3\times 45J=933.4\ J\)

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