Physics

Thermal Properties and Thermodynamics

380 Questions

Thermal properties and thermodynamics questions evaluate concepts of heat transfer, thermal efficiency, and temperature variations. Problems involve calculating heat content, conductivity, and the performance of heat engines. This subject is regularly tested in physics sections across multiple competitive platforms.

Heat transfer calculationsThermal efficiencyBlack body radiationTemperature variationsRefrigeration performance

Thermal Properties and Thermodynamics Questions

Multiple choice 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

For a particular heat engine, 3,000 J of energy goes in at 700 K and 2000 J comes out at 200 K. The rest of the energy is used work.
What is the actual efficiency of this engine?

  1. 0.71

  2. 0.33

  3. 0.67

  4. 0.29

  5. 1.5
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Amount of heat goes in the engine        $Q _H = 3000$ J

Amount of heat rejected by the engine      $Q _R = 2000$ J
Thus work done by the engine      $W = Q _H - Q _R =3000 - 2000 = 1000$ J

Actual efficiency of heat engine       $\eta = \dfrac{W}{Q _H} = \dfrac{1000}{3000} = 0.33$

Multiple choice 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

A 60J of heat is added to the system, resulting in 15J of work being done by the system. The remaining 45J of heat is released. Find out the efficiency of the system?

  1. 100%

  2. 75 %

  3. 45%

  4. 25%

  5. 15%
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Amount of heat absorbed       $Q _H = 60$ J

Net work done by the system     $W= 15$ J
Thus efficiency of the system          $\eta = \dfrac{W}{Q _H} = \dfrac{15}{60}  =0.25$
Thus the system is $25$%  efficient.

Multiple choice 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

An ideal heat engine working between temperatures $T _1$ and $T _2$ has an efficiency $\eta $ . The new efficiency if the temperatures of both the source and sink are doubled, will be 

  1. $\frac{\eta }{2}$
  2. $\eta $
  3. $ 2 \eta $
  4. $ 3 \eta $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Efficiency of heat engine $\eta=\dfrac{output}{input}=\dfrac{{T} _{1}-{T} _{2}}{{T} _{1}}$-------(a)
Now new efficiency of heat engine when sin and source temperature is doubled${\eta} _{new}=\dfrac{2{T} _{1}-2{T} _{2}}{2{T} _{1}}$
${\eta} _{new}=2\dfrac{{T} _{1}-{T} _{2}}{2{T} _{1}}=\eta$ from a
  
Multiple choice 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

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 $W$.

  1. $W = Q _1 - Q _2$
  2. $\displaystyle \eta = \frac{W}{Q _1}$
  3. $\displaystyle \eta = \frac{Q _2}{Q _1}$
  4. $\displaystyle \eta = 1 - \frac{Q _2}{Q _1}$
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
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

Multiple choice 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

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

  1. 0.4

  2. 0.5

  3. 0.6

  4. 0.7
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

Multiple choice 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

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? 

  1. 100%

  2. 70%

  3. 42%

  4. 40%

  5. 30%
Reveal answer Fill a bubble to check yourself
E Correct answer
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, $\eta=\dfrac{30}{100}\times 100=30$ %

Multiple choice 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

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?

  1. 33 percent

  2. 67 percent

  3. 49 percent

  4. 300 percent

  5. 23 percent
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

Multiple choice 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

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

  1. increases

  2. decreases

  3. remains constant

  4. becomes 1
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice 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

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 :

  1. $750\ K$
  2. $500 \ K$
  3. $250\ K$
  4. $1000\ K$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.


Multiple choice 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

An engine has an efficiency of 0.25 when temperature of sink is reduced by 58C,if its efficiency is doubled, then the temperature of the source is:

  1. 150K

  2. 222K

  3. 242K

  4. 232K
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let us consider $T _1=$ temperature of the source

$T _2=$ temperature of the sink
The efficiency of the engine,
$0.25=1-\dfrac{T _2}{T _1}$
$\dfrac{T _2}{T _1}=0.75$
$T _2=0.75T _1$. . . . . . . . .(1)
When the temperature of the sink is reduced by $58^0C$, then the efficiency is double,
$2\times 0.25=1-\dfrac{T _2-58}{T _1}$
$\dfrac{T _2-58}{T _1}=0.5$
$T _2-58=0.5T _1$
$0.75T _1-58=0.5T _1$ .........(from equation 1)
$0.25T _1-58=0$
$0.25T _1=58$
$T _1=\dfrac{58}{0.25}$
$T _1=232 K$
The correct option is D.

Multiple choice physics effects of electric current joule's law of heating thermal effect of electric current heating effect of electric current

A/c to Joules's law, if the potential difference across a conductor having a material of specific resistance (P) remains constant, then the heat produced in the conductor is directly proportional to

  1. $\displaystyle P$
  2. $\displaystyle { P }^{ 2 }$
  3. $\displaystyle \frac { 1 }{ \sqrt { P } } $
  4. $\displaystyle \frac { 1 }{ P } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the specific resistance of the material be  $P$.

$\therefore$  Resistance of the conductor        $R = \dfrac{P L}{A}$
According to Joule's heating effect,  heat produced     $H = \dfrac{V^2t}{R}$

$\therefore$   $H = \dfrac{V^2 t}{PL/A}  = \dfrac{V^2A t}{L}\dfrac{1}{P}$              $\implies H\propto \dfrac{1}{P}$

Multiple choice physics effects of electric current joule's law of heating thermal effect of electric current heating effect of electric current

J (Joule's mechanical equivalent of heat) is equal to

  1. 2400J
  2. $\displaystyle 4.18J{ cal }^{ -1 }$
  3. $\displaystyle 2.2J{ cal }^{ -1 }$

  4. 1000J
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
According to Joule's law, mechanical equivalent of heat    $J = \dfrac{W}{Q}$
Mathematically    $J = 4.18$  $Jcal^{-1}$ if work done and heat produced is measured in joules and calories respectively.
Multiple choice physics effects of electric current joule's law of heating thermal effect of electric current heating effect of electric current

From below which one is the equation for Joule's law of heating effect of electricity?

  1. $H=I^2Rt$
  2. $H=IR^2t$
  3. $H=IRt^2$
  4. $H=\dfrac{It^2}{R}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

According to Joule's law of heating,  heat produced in a resistor     $H = I^2Rt$

where $I$ and $t$ are the current flowing through the resistor and time of current flowing respectively.