148661
The work done by a Carnot engine operating between $300 \mathrm{~K}$ and $400 \mathrm{~K}$ is $400 \mathrm{~J}$. The energy exhausted by the engine is
148662
A Carnot engine operates between heat reservoirs differing in temperature by $80^{\circ} \mathrm{C}$. The efficiency of the Carnot engine is $20 \%$. The temperature of the cold reservoir is
1 $440 \mathrm{~K}$
2 $400 \mathrm{~K}$
3 $250 \mathrm{~K}$
4 $320 \mathrm{~K}$
Explanation:
D Given, $\mathrm{T}_{\mathrm{H}}-\mathrm{T}_{\mathrm{L}}=80 \mathrm{~K}=80^{\circ} \mathrm{C}$ $\eta=20 \% \text { or } \eta=0.2$ Efficiency of Carnot engine, $\eta=1-\frac{T_{L}}{T_{H}}$ $\eta=\frac{T_{H}-T_{L}}{T_{H}}$ $0.2=\frac{80}{T_{H}}$ $T_{H}=\frac{80}{0.2}=400$ $T_{H}=400 \mathrm{~K}$ From equation (i), $\mathrm{T}_{H}-\mathrm{T}_{\mathrm{L}}=80 \mathrm{~K}$ $400-\mathrm{T}_{\mathrm{L}}=80 \mathrm{~K}$ $\mathrm{~T}_{\mathrm{L}}=400-80 \mathrm{~K}$ $\mathrm{~T}_{\mathrm{L}}=320 \mathrm{~K}$
AP EAMCET-11.07.2022
Thermodynamics
148663
A gas in a closed container undergoes the cycle ABCDA as shown in the figure. The net heat absorbed by the gas after it has completed 20 cycles is
1 $+5.0 \mathrm{~kJ}$
2 $-5.0 \mathrm{~kJ}$
3 $+2.5 \mathrm{~kJ}$
4 $-2.5 \mathrm{~kJ}$
Explanation:
A Since we know that area under the curve of P-V diagram gives us work done. So, Work done in one complete cycle $=$ Area under the curve on $\mathrm{P}-\mathrm{V}$ diagram $\mathrm{dW}=\frac{1}{2}[10+15] \times 20$ $\mathrm{dW}=\frac{1}{2}[25 \times 20]=250 \text { Joule }$ Since we know that for a cyclic process net heat transfer is equal to net work transfer. $\quad \oint \mathrm{dQ}=\oint \mathrm{dW}$ $\therefore \quad \mathrm{dQ}=\mathrm{dW}=250 \mathrm{~J} \quad \text { [for one cycle] }$ $\text { And net heat transfer for } 20 \text { cycle, }$ $\mathrm{dQ}=250 \times 20 \mathrm{~J}$ $\mathrm{dQ}=5000 \mathrm{~J}$ $\mathrm{dQ}=+5 \mathrm{~kJ}$
AP EAMCET-11.07.2022
Thermodynamics
148664
A Carnot engine operating between $430 \mathrm{~K}$ and $330 \mathrm{~K}$ does a work $60 \mathrm{~kJ}$. The amount of ice, that can melt from its exhaust is (Latent heat of fusion of ice $=\mathbf{3 3 0 ~} \mathbf{~ g ~ g}^{-\mathbf{1}}$ )
148661
The work done by a Carnot engine operating between $300 \mathrm{~K}$ and $400 \mathrm{~K}$ is $400 \mathrm{~J}$. The energy exhausted by the engine is
148662
A Carnot engine operates between heat reservoirs differing in temperature by $80^{\circ} \mathrm{C}$. The efficiency of the Carnot engine is $20 \%$. The temperature of the cold reservoir is
1 $440 \mathrm{~K}$
2 $400 \mathrm{~K}$
3 $250 \mathrm{~K}$
4 $320 \mathrm{~K}$
Explanation:
D Given, $\mathrm{T}_{\mathrm{H}}-\mathrm{T}_{\mathrm{L}}=80 \mathrm{~K}=80^{\circ} \mathrm{C}$ $\eta=20 \% \text { or } \eta=0.2$ Efficiency of Carnot engine, $\eta=1-\frac{T_{L}}{T_{H}}$ $\eta=\frac{T_{H}-T_{L}}{T_{H}}$ $0.2=\frac{80}{T_{H}}$ $T_{H}=\frac{80}{0.2}=400$ $T_{H}=400 \mathrm{~K}$ From equation (i), $\mathrm{T}_{H}-\mathrm{T}_{\mathrm{L}}=80 \mathrm{~K}$ $400-\mathrm{T}_{\mathrm{L}}=80 \mathrm{~K}$ $\mathrm{~T}_{\mathrm{L}}=400-80 \mathrm{~K}$ $\mathrm{~T}_{\mathrm{L}}=320 \mathrm{~K}$
AP EAMCET-11.07.2022
Thermodynamics
148663
A gas in a closed container undergoes the cycle ABCDA as shown in the figure. The net heat absorbed by the gas after it has completed 20 cycles is
1 $+5.0 \mathrm{~kJ}$
2 $-5.0 \mathrm{~kJ}$
3 $+2.5 \mathrm{~kJ}$
4 $-2.5 \mathrm{~kJ}$
Explanation:
A Since we know that area under the curve of P-V diagram gives us work done. So, Work done in one complete cycle $=$ Area under the curve on $\mathrm{P}-\mathrm{V}$ diagram $\mathrm{dW}=\frac{1}{2}[10+15] \times 20$ $\mathrm{dW}=\frac{1}{2}[25 \times 20]=250 \text { Joule }$ Since we know that for a cyclic process net heat transfer is equal to net work transfer. $\quad \oint \mathrm{dQ}=\oint \mathrm{dW}$ $\therefore \quad \mathrm{dQ}=\mathrm{dW}=250 \mathrm{~J} \quad \text { [for one cycle] }$ $\text { And net heat transfer for } 20 \text { cycle, }$ $\mathrm{dQ}=250 \times 20 \mathrm{~J}$ $\mathrm{dQ}=5000 \mathrm{~J}$ $\mathrm{dQ}=+5 \mathrm{~kJ}$
AP EAMCET-11.07.2022
Thermodynamics
148664
A Carnot engine operating between $430 \mathrm{~K}$ and $330 \mathrm{~K}$ does a work $60 \mathrm{~kJ}$. The amount of ice, that can melt from its exhaust is (Latent heat of fusion of ice $=\mathbf{3 3 0 ~} \mathbf{~ g ~ g}^{-\mathbf{1}}$ )
148661
The work done by a Carnot engine operating between $300 \mathrm{~K}$ and $400 \mathrm{~K}$ is $400 \mathrm{~J}$. The energy exhausted by the engine is
148662
A Carnot engine operates between heat reservoirs differing in temperature by $80^{\circ} \mathrm{C}$. The efficiency of the Carnot engine is $20 \%$. The temperature of the cold reservoir is
1 $440 \mathrm{~K}$
2 $400 \mathrm{~K}$
3 $250 \mathrm{~K}$
4 $320 \mathrm{~K}$
Explanation:
D Given, $\mathrm{T}_{\mathrm{H}}-\mathrm{T}_{\mathrm{L}}=80 \mathrm{~K}=80^{\circ} \mathrm{C}$ $\eta=20 \% \text { or } \eta=0.2$ Efficiency of Carnot engine, $\eta=1-\frac{T_{L}}{T_{H}}$ $\eta=\frac{T_{H}-T_{L}}{T_{H}}$ $0.2=\frac{80}{T_{H}}$ $T_{H}=\frac{80}{0.2}=400$ $T_{H}=400 \mathrm{~K}$ From equation (i), $\mathrm{T}_{H}-\mathrm{T}_{\mathrm{L}}=80 \mathrm{~K}$ $400-\mathrm{T}_{\mathrm{L}}=80 \mathrm{~K}$ $\mathrm{~T}_{\mathrm{L}}=400-80 \mathrm{~K}$ $\mathrm{~T}_{\mathrm{L}}=320 \mathrm{~K}$
AP EAMCET-11.07.2022
Thermodynamics
148663
A gas in a closed container undergoes the cycle ABCDA as shown in the figure. The net heat absorbed by the gas after it has completed 20 cycles is
1 $+5.0 \mathrm{~kJ}$
2 $-5.0 \mathrm{~kJ}$
3 $+2.5 \mathrm{~kJ}$
4 $-2.5 \mathrm{~kJ}$
Explanation:
A Since we know that area under the curve of P-V diagram gives us work done. So, Work done in one complete cycle $=$ Area under the curve on $\mathrm{P}-\mathrm{V}$ diagram $\mathrm{dW}=\frac{1}{2}[10+15] \times 20$ $\mathrm{dW}=\frac{1}{2}[25 \times 20]=250 \text { Joule }$ Since we know that for a cyclic process net heat transfer is equal to net work transfer. $\quad \oint \mathrm{dQ}=\oint \mathrm{dW}$ $\therefore \quad \mathrm{dQ}=\mathrm{dW}=250 \mathrm{~J} \quad \text { [for one cycle] }$ $\text { And net heat transfer for } 20 \text { cycle, }$ $\mathrm{dQ}=250 \times 20 \mathrm{~J}$ $\mathrm{dQ}=5000 \mathrm{~J}$ $\mathrm{dQ}=+5 \mathrm{~kJ}$
AP EAMCET-11.07.2022
Thermodynamics
148664
A Carnot engine operating between $430 \mathrm{~K}$ and $330 \mathrm{~K}$ does a work $60 \mathrm{~kJ}$. The amount of ice, that can melt from its exhaust is (Latent heat of fusion of ice $=\mathbf{3 3 0 ~} \mathbf{~ g ~ g}^{-\mathbf{1}}$ )
148661
The work done by a Carnot engine operating between $300 \mathrm{~K}$ and $400 \mathrm{~K}$ is $400 \mathrm{~J}$. The energy exhausted by the engine is
148662
A Carnot engine operates between heat reservoirs differing in temperature by $80^{\circ} \mathrm{C}$. The efficiency of the Carnot engine is $20 \%$. The temperature of the cold reservoir is
1 $440 \mathrm{~K}$
2 $400 \mathrm{~K}$
3 $250 \mathrm{~K}$
4 $320 \mathrm{~K}$
Explanation:
D Given, $\mathrm{T}_{\mathrm{H}}-\mathrm{T}_{\mathrm{L}}=80 \mathrm{~K}=80^{\circ} \mathrm{C}$ $\eta=20 \% \text { or } \eta=0.2$ Efficiency of Carnot engine, $\eta=1-\frac{T_{L}}{T_{H}}$ $\eta=\frac{T_{H}-T_{L}}{T_{H}}$ $0.2=\frac{80}{T_{H}}$ $T_{H}=\frac{80}{0.2}=400$ $T_{H}=400 \mathrm{~K}$ From equation (i), $\mathrm{T}_{H}-\mathrm{T}_{\mathrm{L}}=80 \mathrm{~K}$ $400-\mathrm{T}_{\mathrm{L}}=80 \mathrm{~K}$ $\mathrm{~T}_{\mathrm{L}}=400-80 \mathrm{~K}$ $\mathrm{~T}_{\mathrm{L}}=320 \mathrm{~K}$
AP EAMCET-11.07.2022
Thermodynamics
148663
A gas in a closed container undergoes the cycle ABCDA as shown in the figure. The net heat absorbed by the gas after it has completed 20 cycles is
1 $+5.0 \mathrm{~kJ}$
2 $-5.0 \mathrm{~kJ}$
3 $+2.5 \mathrm{~kJ}$
4 $-2.5 \mathrm{~kJ}$
Explanation:
A Since we know that area under the curve of P-V diagram gives us work done. So, Work done in one complete cycle $=$ Area under the curve on $\mathrm{P}-\mathrm{V}$ diagram $\mathrm{dW}=\frac{1}{2}[10+15] \times 20$ $\mathrm{dW}=\frac{1}{2}[25 \times 20]=250 \text { Joule }$ Since we know that for a cyclic process net heat transfer is equal to net work transfer. $\quad \oint \mathrm{dQ}=\oint \mathrm{dW}$ $\therefore \quad \mathrm{dQ}=\mathrm{dW}=250 \mathrm{~J} \quad \text { [for one cycle] }$ $\text { And net heat transfer for } 20 \text { cycle, }$ $\mathrm{dQ}=250 \times 20 \mathrm{~J}$ $\mathrm{dQ}=5000 \mathrm{~J}$ $\mathrm{dQ}=+5 \mathrm{~kJ}$
AP EAMCET-11.07.2022
Thermodynamics
148664
A Carnot engine operating between $430 \mathrm{~K}$ and $330 \mathrm{~K}$ does a work $60 \mathrm{~kJ}$. The amount of ice, that can melt from its exhaust is (Latent heat of fusion of ice $=\mathbf{3 3 0 ~} \mathbf{~ g ~ g}^{-\mathbf{1}}$ )
