The steam engine is not a Carnot engine, because all the process involved in the steam engine are not perfectly reversible. But we can calculate the maximum possible efficiency of the steam engine by considering it as a Carnot engine.
\(\eta =1-\frac { { T }_{ L } }{ { T }_{ H } } =1-\frac { 300K }{ 523K } =0.43\)
The steam engine can have maximum possible 43% of efficiency, implying this steam engine can convert 43% of input heat into useful work and remaining 57% is ejected as heat. In practice the efficiency is even less than 43%.

When the water is heated from 30°C to 60°C, there is only a slight change in its volume. So we can treat this process as isochoric. In an isochoric process the work done by the system is zero. The given heat supplied is used to increase only the internal energy.
ΔU = Q = ms_v ΔT
The mass of water = 500 g = 0.5 kg
The change in temperature = 30K
The heat Q = 0.5\(\times\)4184\(\times\)30 = 62.76 kJ

Here, the process is adiabatic compression. The volume is given and temperature is to be found. we can use the equation (8.38 )
\({ T }_{ i }{ V }_{ i }^{ \Upsilon -1 }{ =T }_{ f }{ V }_{ f }^{ \Upsilon -1 }.\)
T_i = 300 K (273 + 27°C = 300 K)
\({ V }_{ i }=V\& { V }_{ f }=\frac { V }{ 4 } \)
\({ T }_{ f }={ T }_{ i }{ \left( \frac { { V }_{ i } }{ { V }_{ f } } \right) }^{ \Upsilon -1 }\) = 300 K × 4^1.4-1 = 300K\(\times\)1.741
T2 ≈ 522 K or 2490C
This temperature is higher than the boiling point of water. So it is very dangerous to touch the nozzle of blocked pump when you pump air.

Work done by the system (body),
W = +500 kJ
Heat released from the system (body),
Q = –230 kJ
The change in internal energy of a body
= \(\Delta\)U= – 230 kJ – 500 kJ = – 730 kJ

(a) We know that work done by the gas in an isothermal expansion
Since μ = 0.5
W = 0.5mol \(\times\)\(\frac { 8.31J }{ mol.K } \times 300 K\ In\ \left( \frac { 6L }{ 2L } \right) \)
W = 1.369 kJ
Note that W is positive since the work is done by the gas.
(b) From the First law of thermodynamics, in an isothermal process the heat supplied is spent to do work.Therefore, Q = W = 1.369 kJ. Thus Q is also positive which implies that heat flows into the system.
(c) For an isothermal process
P_iV_i = P_fV_f = μRT
\({ P }_{ f }=\frac { \mu RT }{ { V }_{ f } } =0.5mol\times \frac { 8.31J }{ mol.K } \times \frac { 300K }{ 6\times { 10 }^{ -3 }{ m }^{ 3 } } \)
= 207.75 k Pa