The source and sink temperatures of a Carnot engine are 400 K and 300 K, respectively. What is its efficiency?
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Solution
Efficiency, \(\eta =1-\frac{T_{2}}{T_{1}}\)
T1(source temp.) = 400 K
T2(sinktemp.) = 300 K
\(∴\eta =1-\frac{300}{400}=\frac{1}{4}=25\)%
The volume of a gas is reduced adiabatically to 1/4 of its volume at 27°C. If γ = 1.4 the new temperature is
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Solution
\((300)V^{(1.4-1)}=T_{2}\cdot \left ( \frac{V}{4} \right )^{(1.4-1)}\)
T2=(300) (4)0.4
A Carnot’s engine works as a refrigerator between 250 K and 300 K. If it receives 750 calories of heat from the reservoir at the lower temperature, the amount of heat rejected at the higher temperature is
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Solution
\(\frac{750}{W}=\frac{250}{300-250}\)
Heat rejected = 750 + 150 = 900 cal.
The efficiency of carnot engine when source temperature is T1 and sink temperature is T2 will be
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Solution
Efficiency of carnot engine\(\eta =\frac{T_{1}-T_{2}}{T_{1}}\)
where T1 =source temperature
T2 =sink temperature.
In the equation PVγ= constant,the value of γ is unity. Then the process is
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Solution
PV = constant represents isothermal process.
For adiabatic processes (Letters have usual meanings)
The gas law \(\frac{PV}{T}\) constant is true for
When heat is given to a gas in an isothermal change, the result will be
A Carnot engine is working between 127°C and 27°C. The increase in efficiency will be maximum when the temperature of
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Solution
\(\frac{T_{1}-T_{2}}{T_{1}}\)is maximum in case (b).
In changing the state of the rmodynamics from A to B state,the heat required is Q and the work done by the system is W.The change in its internal energy is
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Solution
ΔQ = ΔU + ΔW
⇒ ΔU = ΔQ - ΔW = Q - W(using proper sign)