Tag: refrigerators and heat pumps

Questions Related to refrigerators and heat pumps

A car is moving with a speed of $40 $ km/hr. If the car engine generated 7 kilowatt power, then the resistive force in the path of the car will be:-
physics heat engine: second law of thermodynamics conversion of heat into work: heat engine and it's efficiency engines and cycles heat engines refrigerators and heat pumps

  1. 360 Netwon

  2. 630 Newton

  3. Zero

  4. 280 Newton

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Correct Option: B
Explanation:

Power= Force $\times$ speed

$R= Power/Speed$
Given, $P=7Kw=7000W$
Speed= $40km/h=11.11m/s$
Resistance= $7000/11.11= 630.063 N$

A system undergoes a cyclic process in which it absorbs $Q _1$ heat and gives out $Q _2$ heat. The efficiency of the process is $\eta$ and the work done is $W$.

physics heat engine: second law of thermodynamics conversion of heat into work: heat engine and it's efficiency engines and cycles heat engines refrigerators and heat pumps

  1. $W = Q _1 - Q _2$

  2. $\displaystyle \eta = \frac{W}{Q _1}$

  3. $\displaystyle \eta = \frac{Q _2}{Q _1}$

  4. $\displaystyle \eta = 1 - \frac{Q _2}{Q _1}$

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Correct Option: A,B,D
Explanation:

The efficiency of a heat engine is defined as the ratio of work output to the heat input.
Thus, $ \eta = \dfrac{W}{{Q} _{in}} $
Now, since the heat energy only gets converted into work and no other form of energy, thus following the law of conservation of energy,
$ W = {Q} _{out} - {Q} _{in} $
Substituting this value in the above expression, we get another expression for the efficiency of a heat engine

A motor pump is delilvering water at certain rate.In order to increase the rate of delivery by100% ,the power of the motor is to be increased by

physics heat engine: second law of thermodynamics conversion of heat into work: heat engine and it's efficiency engines and cycles heat engines refrigerators and heat pumps

  1. 300%

  2. 200%

  3. 400%

  4. 700%

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Correct Option: A

The efficiency of a heat engine if the temperature of source $227^0C$ and that of sink is $27^0C$ nearly

physics heat engine: second law of thermodynamics conversion of heat into work: heat engine and it's efficiency engines and cycles heat engines refrigerators and heat pumps

  1. 0.4

  2. 0.5

  3. 0.6

  4. 0.7

Show answer
Correct Option: C
Explanation:

The efficiency of heat engine will be$:$

${T _{\sin k}}/{T _{source}}$
$ = \left( {273 + 27} \right)/\left( {273 + 227} \right)$
$ = 300/500$
$ = 0.6$
So$,$ efficiency will be $0.6$
Hence,
option $(C)$ is correct answer.

A heat engine produces 100 J of heat, does 30 J of work, and emits 70 J into a cold reservoir. What is the efficiency of the heat engine? 

physics heat engine: second law of thermodynamics conversion of heat into work: heat engine and it's efficiency engines and cycles heat engines refrigerators and heat pumps

  1. 100%

  2. 70%

  3. 42%

  4. 40%

  5. 30%

Show answer
Correct Option: E
Explanation:

The efficiency of heat engine is $\eta=\dfrac{W}{Q _{in}}$  

Here work done , $W=30 J$ and heat produced $Q _{in}=100 J$
Thus, % of efficiency, $\eta=\dfrac{30}{100}\times 100=30$ %

A heat engine takes in heat at 750 degrees Celsius and expels heat at 250 degrees Celsius. What is this engine's theoretically ideal (Carnot) efficiency?

physics heat engine: second law of thermodynamics conversion of heat into work: heat engine and it's efficiency engines and cycles heat engines refrigerators and heat pumps

  1. 33 percent

  2. 67 percent

  3. 49 percent

  4. 300 percent

  5. 23 percent

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Correct Option: C
Explanation:

Temperature of source     $T _H = 750^oC = 750+273 = 1023$  K

Temperature of sink     $T _L = 250^oC = 250+273 = 523$  K

Efficiency of engine        $\eta = 1-\dfrac{T _L}{T _H} = 1  -\dfrac{523}{1023}  =0.49$
Thus the engine is  $49$% efficient.

Let $E _D, E _P, E _S$ denote efficiency of a diesel, a petrol and a steam engine respectively. which of the following is correct?

physics heat engine: second law of thermodynamics conversion of heat into work: heat engine and it's efficiency engines and cycles heat engines refrigerators and heat pumps

  1. $E _D < E _P > E _S$

  2. $E _D < E _P < E _S$

  3. $E _P > E _D > E _S$

  4. $E _P < E _S < E _D$

Show answer
Correct Option: C
Explanation:

Let, $E _D,E _P,E _S$ denote efficiency of a diesel, a petrol stream engine respective.

$E _P>E _D>E _S$
because petrol and diesel using the Carnot cycle.

NA heat engine has an efficiency n.Temperatures of source and sink are each decreased by 100 K. The efficiency of the engine:

physics heat engine: second law of thermodynamics conversion of heat into work: heat engine and it's efficiency engines and cycles heat engines refrigerators and heat pumps

  1. increases

  2. decreases

  3. remains constant

  4. becomes 1

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Correct Option: A
Explanation:

Efficiency of heat engine,

$n=1-\dfrac{T _2}{T _1}=\dfrac{T _1-T _2}{T _1}$. . . . . .(1)
According to question,
 $T _1\rightarrow T _1-100$
$T _2\rightarrow T _2-100$
New efficiency, $n'=1-\dfrac{T _2-100}{T _1-100}$
$n'=\dfrac{T _1-T _2}{T _1-100}$
From equation (1) and (2), we can conclude that
$n'\propto \dfrac{1}{T _1-100}$
The efficiency of the engine is increases.
The correct option is A.

The most efficient engine is?

physics heat engine: second law of thermodynamics conversion of heat into work: heat engine and it's efficiency engines and cycles heat engines refrigerators and heat pumps

  1. Petrol

  2. Electric

  3. Steam

  4. Diesel

Show answer
Correct Option: B
Explanation:

The electric engine is most efficient. Electric motors are very efficient at converting electricity into work. Physicists throw around abstract numbers like 90%, whereas when discussing combustion engines, they use numbers less than 40%. But you cannot compare the efficiency of converting electricity into motion to the efficiency of converting gasoline into motion.

Efficiency of a heat engine whose sink is at a temperature of $300 \ K$ is $40$%. To increase the efficiency to $60$%, keeping the sink temperature constant, the source temperature must be increased by :

physics heat engine: second law of thermodynamics conversion of heat into work: heat engine and it's efficiency engines and cycles heat engines refrigerators and heat pumps

  1. $750\  K$

  2. $500 \ K$

  3. $250\  K$

  4. $1000\  K$

Show answer
Correct Option: C
Explanation:

We know that in heat engine, sink is at lower temperature than source, hence,


$\eta =1-\dfrac { { T } _{ sink } }{ { T } _{ source } } $

$0.4=1-\dfrac { 300 }{ { T } _{ source } } $

${ T } _{ source } = 500\ K$

In the second case,

$0.6=1-\dfrac { 300 }{ { T } _{ source } } $

${ T } _{ source } = 750\ K$

Hence, an increment of $250\ K$ in source temperature is required.