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PHXI12:THERMODYNAMICS

371358 A Carnot engine takes 300 calories of heat from a source at $500 K$ and rejects 150 calories of heat to the sink. The temperature of the sink is

1

125 K

2

250 K

3

750 K

4

1000 K

Explanation:

Temperature of the source, $T_{1} = 500 K$
Heat taken from the source, $Q_{1} = 300 c a l$
Heat rejected to the sink, $Q_{2} = 150 c a l$
Temperature of the sink, $T_{2} =$ ?
As $\frac{T_{1}}{T_{2}} = \frac{Q_{1}}{Q_{2}} ∴ T_{2} = \frac{Q_{2}}{Q_{1}} T_{1}$
or $T_{2} = (\frac{150 c a l}{300 c a l}) (500 K) = 250 K$

KCET - 2018

PHXI12:THERMODYNAMICS

371359 A Carnot engine takes heat from a reservoir at $627^{\circ} C$ and rejects heat to a sink at $27^{\circ} C$ . Its efficiency will be

1

3 / 5

2

1 / 3

3

2 / 3

4

200 / 209

Explanation:

Efficiency of carnot engine,
$\begin{aligned} η = 1 - \frac{T_{2}}{T_{1}} \\ η = 1 - \frac{27 + 273}{627 + 273} = 1 - \frac{300}{900} = \frac{600}{900} = \frac{2}{3} \end{aligned}$

KCET - 2006

PHXI12:THERMODYNAMICS

371360 If the energy input to a Carnot engine is thrice the work it performs then, the fraction of energy rejected to the sink is

1

\frac{1}{3}

2

\frac{1}{4}

3

\frac{2}{5}

4

\frac{2}{3}

Explanation:

Efficiency $η = \frac{W_{ouput}}{{Heat}_{input}} = \frac{w}{3 w} = \frac{1}{3}$
$η = 1 - \frac{Q_{2}}{Q_{1}} = \frac{1}{3} ∴ \frac{Q_{2}}{Q_{1}} = \frac{2}{3}$

PHXI12:THERMODYNAMICS

371361 A Carnot engine absorbs $6 \times 10^{5} c a l$ at $227^{\circ} C$ The work done per cycle by the engine, if its sink is maintained at $127^{\circ} C$ is

1

15 \times 10^{8} J

2

15 \times 10^{4} J

3

5 \times 10^{5} J

4

2 \times 10^{4} J

Explanation:

Here, $Q_{1} = 6 \times 10^{5} cal$
$T_{1} = 227^{\circ} C = 227 + 273 = 500 K$
$T_{2} = 127^{\circ} C = 127 + 273 = 400 K$
Work done/cycle, $W =$ ?
As, $\frac{Q_{2}}{Q_{1}} = \frac{T_{2}}{T_{1}} o r Q_{2} = \frac{T_{2}}{T_{1}} \times Q_{1}$
$= \frac{400}{500} \times 6 \times 10^{5}$
$= 4.8 \times 10^{5} c a l$
As, $W = Q_{1} - Q_{2} = 6 \times 10^{5} - 4.8 \times 10^{5}$
$(∵ 1 c a l = 4.2 J)$
$= 1.2 \times 10^{5} \times 4.2 J$
$\Rightarrow W = 5.04 \times 10^{5} J \approx 5 \times 10^{5} J$

PHXI12:THERMODYNAMICS

371362 Which of the following laws of thermodynamics leads to the inference that it is difficult to convert whole of heat into work

1 zeroth

2 second

3 first

4 third

PHXI12:THERMODYNAMICS

371358 A Carnot engine takes 300 calories of heat from a source at $500 K$ and rejects 150 calories of heat to the sink. The temperature of the sink is

1

125 K

2

250 K

3

750 K

4

1000 K

Explanation:

Temperature of the source, $T_{1} = 500 K$
Heat taken from the source, $Q_{1} = 300 c a l$
Heat rejected to the sink, $Q_{2} = 150 c a l$
Temperature of the sink, $T_{2} =$ ?
As $\frac{T_{1}}{T_{2}} = \frac{Q_{1}}{Q_{2}} ∴ T_{2} = \frac{Q_{2}}{Q_{1}} T_{1}$
or $T_{2} = (\frac{150 c a l}{300 c a l}) (500 K) = 250 K$

KCET - 2018

PHXI12:THERMODYNAMICS

371359 A Carnot engine takes heat from a reservoir at $627^{\circ} C$ and rejects heat to a sink at $27^{\circ} C$ . Its efficiency will be

1

3 / 5

2

1 / 3

3

2 / 3

4

200 / 209

Explanation:

Efficiency of carnot engine,
$\begin{aligned} η = 1 - \frac{T_{2}}{T_{1}} \\ η = 1 - \frac{27 + 273}{627 + 273} = 1 - \frac{300}{900} = \frac{600}{900} = \frac{2}{3} \end{aligned}$

KCET - 2006

PHXI12:THERMODYNAMICS

371360 If the energy input to a Carnot engine is thrice the work it performs then, the fraction of energy rejected to the sink is

1

\frac{1}{3}

2

\frac{1}{4}

3

\frac{2}{5}

4

\frac{2}{3}

Explanation:

Efficiency $η = \frac{W_{ouput}}{{Heat}_{input}} = \frac{w}{3 w} = \frac{1}{3}$
$η = 1 - \frac{Q_{2}}{Q_{1}} = \frac{1}{3} ∴ \frac{Q_{2}}{Q_{1}} = \frac{2}{3}$

PHXI12:THERMODYNAMICS

371361 A Carnot engine absorbs $6 \times 10^{5} c a l$ at $227^{\circ} C$ The work done per cycle by the engine, if its sink is maintained at $127^{\circ} C$ is

1

15 \times 10^{8} J

2

15 \times 10^{4} J

3

5 \times 10^{5} J

4

2 \times 10^{4} J

Explanation:

Here, $Q_{1} = 6 \times 10^{5} cal$
$T_{1} = 227^{\circ} C = 227 + 273 = 500 K$
$T_{2} = 127^{\circ} C = 127 + 273 = 400 K$
Work done/cycle, $W =$ ?
As, $\frac{Q_{2}}{Q_{1}} = \frac{T_{2}}{T_{1}} o r Q_{2} = \frac{T_{2}}{T_{1}} \times Q_{1}$
$= \frac{400}{500} \times 6 \times 10^{5}$
$= 4.8 \times 10^{5} c a l$
As, $W = Q_{1} - Q_{2} = 6 \times 10^{5} - 4.8 \times 10^{5}$
$(∵ 1 c a l = 4.2 J)$
$= 1.2 \times 10^{5} \times 4.2 J$
$\Rightarrow W = 5.04 \times 10^{5} J \approx 5 \times 10^{5} J$

PHXI12:THERMODYNAMICS

371362 Which of the following laws of thermodynamics leads to the inference that it is difficult to convert whole of heat into work

1 zeroth

2 second

3 first

4 third

PHXI12:THERMODYNAMICS

371358 A Carnot engine takes 300 calories of heat from a source at $500 K$ and rejects 150 calories of heat to the sink. The temperature of the sink is

1

125 K

2

250 K

3

750 K

4

1000 K

Explanation:

Temperature of the source, $T_{1} = 500 K$
Heat taken from the source, $Q_{1} = 300 c a l$
Heat rejected to the sink, $Q_{2} = 150 c a l$
Temperature of the sink, $T_{2} =$ ?
As $\frac{T_{1}}{T_{2}} = \frac{Q_{1}}{Q_{2}} ∴ T_{2} = \frac{Q_{2}}{Q_{1}} T_{1}$
or $T_{2} = (\frac{150 c a l}{300 c a l}) (500 K) = 250 K$

KCET - 2018

PHXI12:THERMODYNAMICS

371359 A Carnot engine takes heat from a reservoir at $627^{\circ} C$ and rejects heat to a sink at $27^{\circ} C$ . Its efficiency will be

1

3 / 5

2

1 / 3

3

2 / 3

4

200 / 209

Explanation:

Efficiency of carnot engine,
$\begin{aligned} η = 1 - \frac{T_{2}}{T_{1}} \\ η = 1 - \frac{27 + 273}{627 + 273} = 1 - \frac{300}{900} = \frac{600}{900} = \frac{2}{3} \end{aligned}$

KCET - 2006

PHXI12:THERMODYNAMICS

371360 If the energy input to a Carnot engine is thrice the work it performs then, the fraction of energy rejected to the sink is

1

\frac{1}{3}

2

\frac{1}{4}

3

\frac{2}{5}

4

\frac{2}{3}

Explanation:

Efficiency $η = \frac{W_{ouput}}{{Heat}_{input}} = \frac{w}{3 w} = \frac{1}{3}$
$η = 1 - \frac{Q_{2}}{Q_{1}} = \frac{1}{3} ∴ \frac{Q_{2}}{Q_{1}} = \frac{2}{3}$

PHXI12:THERMODYNAMICS

371361 A Carnot engine absorbs $6 \times 10^{5} c a l$ at $227^{\circ} C$ The work done per cycle by the engine, if its sink is maintained at $127^{\circ} C$ is

1

15 \times 10^{8} J

2

15 \times 10^{4} J

3

5 \times 10^{5} J

4

2 \times 10^{4} J

Explanation:

Here, $Q_{1} = 6 \times 10^{5} cal$
$T_{1} = 227^{\circ} C = 227 + 273 = 500 K$
$T_{2} = 127^{\circ} C = 127 + 273 = 400 K$
Work done/cycle, $W =$ ?
As, $\frac{Q_{2}}{Q_{1}} = \frac{T_{2}}{T_{1}} o r Q_{2} = \frac{T_{2}}{T_{1}} \times Q_{1}$
$= \frac{400}{500} \times 6 \times 10^{5}$
$= 4.8 \times 10^{5} c a l$
As, $W = Q_{1} - Q_{2} = 6 \times 10^{5} - 4.8 \times 10^{5}$
$(∵ 1 c a l = 4.2 J)$
$= 1.2 \times 10^{5} \times 4.2 J$
$\Rightarrow W = 5.04 \times 10^{5} J \approx 5 \times 10^{5} J$

PHXI12:THERMODYNAMICS

371362 Which of the following laws of thermodynamics leads to the inference that it is difficult to convert whole of heat into work

1 zeroth

2 second

3 first

4 third

PHXI12:THERMODYNAMICS

371358 A Carnot engine takes 300 calories of heat from a source at $500 K$ and rejects 150 calories of heat to the sink. The temperature of the sink is

1

125 K

2

250 K

3

750 K

4

1000 K

Explanation:

Temperature of the source, $T_{1} = 500 K$
Heat taken from the source, $Q_{1} = 300 c a l$
Heat rejected to the sink, $Q_{2} = 150 c a l$
Temperature of the sink, $T_{2} =$ ?
As $\frac{T_{1}}{T_{2}} = \frac{Q_{1}}{Q_{2}} ∴ T_{2} = \frac{Q_{2}}{Q_{1}} T_{1}$
or $T_{2} = (\frac{150 c a l}{300 c a l}) (500 K) = 250 K$

KCET - 2018

PHXI12:THERMODYNAMICS

371359 A Carnot engine takes heat from a reservoir at $627^{\circ} C$ and rejects heat to a sink at $27^{\circ} C$ . Its efficiency will be

1

3 / 5

2

1 / 3

3

2 / 3

4

200 / 209

Explanation:

Efficiency of carnot engine,
$\begin{aligned} η = 1 - \frac{T_{2}}{T_{1}} \\ η = 1 - \frac{27 + 273}{627 + 273} = 1 - \frac{300}{900} = \frac{600}{900} = \frac{2}{3} \end{aligned}$

KCET - 2006

PHXI12:THERMODYNAMICS

371360 If the energy input to a Carnot engine is thrice the work it performs then, the fraction of energy rejected to the sink is

1

\frac{1}{3}

2

\frac{1}{4}

3

\frac{2}{5}

4

\frac{2}{3}

Explanation:

Efficiency $η = \frac{W_{ouput}}{{Heat}_{input}} = \frac{w}{3 w} = \frac{1}{3}$
$η = 1 - \frac{Q_{2}}{Q_{1}} = \frac{1}{3} ∴ \frac{Q_{2}}{Q_{1}} = \frac{2}{3}$

PHXI12:THERMODYNAMICS

371361 A Carnot engine absorbs $6 \times 10^{5} c a l$ at $227^{\circ} C$ The work done per cycle by the engine, if its sink is maintained at $127^{\circ} C$ is

1

15 \times 10^{8} J

2

15 \times 10^{4} J

3

5 \times 10^{5} J

4

2 \times 10^{4} J

Explanation:

Here, $Q_{1} = 6 \times 10^{5} cal$
$T_{1} = 227^{\circ} C = 227 + 273 = 500 K$
$T_{2} = 127^{\circ} C = 127 + 273 = 400 K$
Work done/cycle, $W =$ ?
As, $\frac{Q_{2}}{Q_{1}} = \frac{T_{2}}{T_{1}} o r Q_{2} = \frac{T_{2}}{T_{1}} \times Q_{1}$
$= \frac{400}{500} \times 6 \times 10^{5}$
$= 4.8 \times 10^{5} c a l$
As, $W = Q_{1} - Q_{2} = 6 \times 10^{5} - 4.8 \times 10^{5}$
$(∵ 1 c a l = 4.2 J)$
$= 1.2 \times 10^{5} \times 4.2 J$
$\Rightarrow W = 5.04 \times 10^{5} J \approx 5 \times 10^{5} J$

PHXI12:THERMODYNAMICS

371362 Which of the following laws of thermodynamics leads to the inference that it is difficult to convert whole of heat into work

1 zeroth

2 second

3 first

4 third

PHXI12:THERMODYNAMICS

371358 A Carnot engine takes 300 calories of heat from a source at $500 K$ and rejects 150 calories of heat to the sink. The temperature of the sink is

1

125 K

2

250 K

3

750 K

4

1000 K

Explanation:

Temperature of the source, $T_{1} = 500 K$
Heat taken from the source, $Q_{1} = 300 c a l$
Heat rejected to the sink, $Q_{2} = 150 c a l$
Temperature of the sink, $T_{2} =$ ?
As $\frac{T_{1}}{T_{2}} = \frac{Q_{1}}{Q_{2}} ∴ T_{2} = \frac{Q_{2}}{Q_{1}} T_{1}$
or $T_{2} = (\frac{150 c a l}{300 c a l}) (500 K) = 250 K$

KCET - 2018

PHXI12:THERMODYNAMICS

371359 A Carnot engine takes heat from a reservoir at $627^{\circ} C$ and rejects heat to a sink at $27^{\circ} C$ . Its efficiency will be

1

3 / 5

2

1 / 3

3

2 / 3

4

200 / 209

Explanation:

Efficiency of carnot engine,
$\begin{aligned} η = 1 - \frac{T_{2}}{T_{1}} \\ η = 1 - \frac{27 + 273}{627 + 273} = 1 - \frac{300}{900} = \frac{600}{900} = \frac{2}{3} \end{aligned}$

KCET - 2006

PHXI12:THERMODYNAMICS

371360 If the energy input to a Carnot engine is thrice the work it performs then, the fraction of energy rejected to the sink is

1

\frac{1}{3}

2

\frac{1}{4}

3

\frac{2}{5}

4

\frac{2}{3}

Explanation:

Efficiency $η = \frac{W_{ouput}}{{Heat}_{input}} = \frac{w}{3 w} = \frac{1}{3}$
$η = 1 - \frac{Q_{2}}{Q_{1}} = \frac{1}{3} ∴ \frac{Q_{2}}{Q_{1}} = \frac{2}{3}$

PHXI12:THERMODYNAMICS

371361 A Carnot engine absorbs $6 \times 10^{5} c a l$ at $227^{\circ} C$ The work done per cycle by the engine, if its sink is maintained at $127^{\circ} C$ is

1

15 \times 10^{8} J

2

15 \times 10^{4} J

3

5 \times 10^{5} J

4

2 \times 10^{4} J

Explanation:

Here, $Q_{1} = 6 \times 10^{5} cal$
$T_{1} = 227^{\circ} C = 227 + 273 = 500 K$
$T_{2} = 127^{\circ} C = 127 + 273 = 400 K$
Work done/cycle, $W =$ ?
As, $\frac{Q_{2}}{Q_{1}} = \frac{T_{2}}{T_{1}} o r Q_{2} = \frac{T_{2}}{T_{1}} \times Q_{1}$
$= \frac{400}{500} \times 6 \times 10^{5}$
$= 4.8 \times 10^{5} c a l$
As, $W = Q_{1} - Q_{2} = 6 \times 10^{5} - 4.8 \times 10^{5}$
$(∵ 1 c a l = 4.2 J)$
$= 1.2 \times 10^{5} \times 4.2 J$
$\Rightarrow W = 5.04 \times 10^{5} J \approx 5 \times 10^{5} J$

PHXI12:THERMODYNAMICS

371362 Which of the following laws of thermodynamics leads to the inference that it is difficult to convert whole of heat into work

1 zeroth

2 second

3 first

4 third