Physics

Thermal Properties and Thermodynamics

431 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 option b: engineering physics conversion of heat into work: heat engine and it's efficiency engines and cycles refrigerators and heat pumps

A reversible engine operates between temperatures 900 K & $T _2$($T _2$ < 900 K), & another reversible engine between $T _2$ & 400 K ($T _2$ > 400 K) in series. What is the value of $T _2$ if work outputs of both the engines are equal?

  1. 600K

  2. 625K

  3. 650K

  4. 675K
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Output work  $W = \Delta T$
Thus for equal work output, temperature difference should be equal.
$900-T _2=T _2-400$
Or  $2T _2 = 1300$ 
$\implies$ $T _2=650 \ K$
Multiple choice physics option b: engineering physics conversion of heat into work: heat engine and it's efficiency engines and cycles refrigerators and heat pumps

A series combination of two Carnots engines operate between the temperatures of $180^0C$ and $20^0C$. If the engines produce equal amount of work,then what is the intermediate temperature(In $^0C$)?

  1. 80

  2. 90

  3. 100

  4. 110
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation


A series combination of two Carnot engines operate between the temperatures of $180^0C$and $20^0C$.If the engines produce equal amount of work
The intermediate temperature in series combination is given by 
$T _i=\dfrac{T _1+T _2}{2}=\dfrac{180+20}{2}=100^oC$
Multiple choice physics option b: engineering physics conversion of heat into work: heat engine and it's efficiency engines and cycles refrigerators and heat pumps

In a cyclic heat engine operating between a source temperature of $600^0C$ and a sink temperature of $20^0 C$, the least rate of heat rejection per kW net output of the engine is,

  1. 0.505kW

  2. 0.490kW

  3. 0.333kW

  4. none of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\therefore Heat\quad absorption=\dfrac { { T } _{ 1 }-{ T } _{ 2 } }{ { T } _{ 1 } } $


$=\dfrac { 873-293 }{ 873 } =0.664$


$\therefore Heat\quad Rejected=1-heat\quad absorbed$
$=1-0.664$
$\therefore Heat\quad rejected=0.335$

Hence the heat of rejection per kW is $0.335$

Multiple choice physics option b: engineering physics conversion of heat into work: heat engine and it's efficiency engines and cycles refrigerators and heat pumps

A heat engine is supplied with 250 kJ/s of heat at a constant fixed temperature of $227^0C$; the heat is rejected at $27^0C$, the cycle is reversible, then what amount of heat is rejected?

  1. 24kJ/s

  2. 223kJ/s

  3. 150kJ/s

  4. none of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The temperature in kelvin scales are $T _1=273+27=300K,T _2=273+237=500K$

$T _1$is temperature of sink And$T _2$ is temperature of source hence by efficiency we get

$\eta=1-\dfrac{T _1}{T _2}=1-\dfrac{Q _1}{Q _2}$


$Q _1=\dfrac{T _1}{T _2}*Q _2=\dfrac{300}{500}*250=150kW$
Multiple choice physics option b: engineering physics conversion of heat into work: heat engine and it's efficiency engines and cycles refrigerators and heat pumps

An engine working on Carnot cycle rejects 40% of absorbed heat from the source, while the sink temperature is maintained at $27^0C$, then what is the source temperature (in $^0C$)?

  1. 477

  2. 346

  3. 564

  4. none of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\eta =\dfrac { 60 }{ 100 } $

${ T } _{ 2 }={ 27 }^{ o }C$
${ T } _{ 2 }={ 300 }^{ o }K$
${ T } _{ 1 }=?$

$\eta =\dfrac { { T } _{ 1 }-{ T } _{ 2 } }{ { T } _{ 1 } } $

Here, ${ T } _{ 1 }=Source\quad temperature$
${ T } _{ 2 }=sink\quad temperature$
$\eta =Efficiency$
$\eta =100-rejecion$

$\therefore \dfrac { 60 }{ 100 } =\dfrac { { T } _{ 1 }-{ T } _{ 2 } }{ { T } _{ 1 } } $

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

$\therefore { T } _{ 1 }\left( 1-0.6 \right) =300$

${ T } _{ 1 }=\dfrac { 300 }{ 0.4 } =\dfrac { 3000 }{ 4 } ={ 750 }^{ o }K$

$\therefore { T } _{ 1 }={ \left( 750-273 \right)  }^{ o }C$
$\therefore { T } _{ 1 }={ 477 }^{ o }C$
Multiple choice physics option b: engineering physics conversion of heat into work: heat engine and it's efficiency engines and cycles refrigerators and heat pumps

One reversible heat engine operates between 1600 K and $T _2$ K, and another reversible heat engine operates between $T _2$ K and 400 K. If both the engines have the same heat input and output, then the temperature $T _2$ must be equal to:

  1. 600

  2. 800

  3. 650

  4. 675
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

${ T } _{ 1 }=1600K$


${ T } _{ 2 }={ T } _{ 2 }K$


${ T } _{ 1 }^{ 1 }={ T } _{ 2 }K$

${ T } _{ 2 }^{ 1 }=400K$

Same input and same output.Then the efficiency is same.
$\eta =\dfrac { { T } _{ 1 }-{ T } _{ 2 } }{ { T } _{ 1 } } $

$\therefore \dfrac { 1600-{ T } _{ 2 } }{ 1600 } =\dfrac { { T } _{ 2 }-400 }{ { T } _{ 2 } } $

$=1600{ T } _{ 2 }-{ T } _{ 2 }^{ 2 }=1600{ T } _{ 2 }-640000$

$\therefore { T } _{ 2 }^{ 2 }=640000$
$\therefore { T } _{ 2 }=800K$

Multiple choice physics thermal physics heat and heat transfer transfer of heat fundamentals of heat transfer

There are two lead spheres, the ratio c being $1 : 2$. If both are at the same tempe then ratio of heat contents is

  1. $1 : 1$
  2. $1 : 2$
  3. $1 : 4$
  4. $1 : 8$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know that the heat content is proportional to the mass and specific heat capacity of the substance as well as temperature. As both the spheres are of the same material and are at the same temperature, heat content will be dependent on mass only.

$ m = \rho \times v$

where $ \rho $ is the density of the sphere and $v$ the volume.

And $ v $ = $ \dfrac {4 \pi r^3}{3} $

The given radius ratio is $ 1:2 $.

Hence heat contents ratio is in the ratio $1:8$

Multiple choice physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

How much heat is required to raise the temperature of 150 g of iron from $20 ^oC$ to $25 ^oC$? (Specific heat of iron $480 J kg^{-1} {\;}^oC^{-1})$

  1. 350 J

  2. 345 J

  3. 360 J

  4. 330 J
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given $m=150$
$g=\frac {150}{1000}=0.15 kg$
Specific heat of iron
$C=480 J kg^{-1} {\;}^oC^{-1})$
$\Delta =(25-20)^oC=5^oC$
$Q=m\times C\times \Delta T$
$=0.15 kg\times 480 J kg^{-1} {\;}^oC^{-1}\times 5^oC$
$=360 J$.

Multiple choice physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

Find the heat lost by a copper cube of mass 400 g when it cool from $100^oC$ to $30^oC$. (Specific of heat of copper $=390 J kg^{-1} {\;}^oC^{-1})$.

  1. 50000 J

  2. 10000 J

  3. 10920 J

  4. 10900 J
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$m=400 g=0.4 kg$
$C=390 J kg^{-1} {\;}^oC^{-1}$
$T _1=100 ^oC, T _2=30^oC$
$\Delta T=70^oC$
$\therefore$ Heat lost $=mC(T _1-T _2)$
$=0.4\times 390\times 70 J=10,920 J$

Multiple choice physics thermal properties heat exchange calorimeter measuring thermal quantities by the method of mixtures

What is the process in which heat energy of both hot and cold body equalizes?

  1. Calorimetry

  2. Fermentation

  3. Latent heat

  4. Hidden heat
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Principle of calorimetry states that heat lost by a hotter body = heat gained by a colder body, therefore calorimetry is a process in which heat energy of both hot and cold body equalizes.

Multiple choice physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

According to principle of calorimetry, heat absorbed by cold bodies is equal to heat released by hot bodies.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The statement is true, according to principle of calorimetry, heat absorbed by cold bodies is equal to heat released by hot bodies.
It can also be understood by conservation of energy concept.
Multiple choice physics calorimetry heat exchange calorimeter measuring thermal quantities by the method of mixtures

$5$g of copper was heated from $20^{\circ}$ to $80^{\circ}$. How much energy was used to heat Cu? (Specific heat capacity of Cu is $0.092 cal/g ^{\circ}C$).

  1. $27.6$ cal
  2. $50$ cal
  3. $35$ cal
  4. $25.7$ cal
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given,

Mass, $m=5\,g$

Specific heat capacity, $C=0.092\,cal/g \,^0C$

Change in temperature, $\Delta T=80\,^0C -20^0C=600^0C$

Heat required, $Q=?$

We have the equation,

$Q=m\times C\times \Delta T$

Then,

$Q=5\times 0.092\times 60=27.6\,cal$
Multiple choice physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

What is the principle of the method of a mixture? Name the law on which this principle is based.

  1. Newtons law of cooling

  2. none

  3. principle of calorimetry

  4. principle of heat transfer
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A principle of calorimetry states that if there is no loss of heat in surrounding the total heat loss by hot body equals to total heat gained by a cold body.
i.e. heat loss = heat gained
It is based on Newton's Law of Cooling, which states that when a liquid is heated of higher temperature and placed to cool. Then the rate of heat lost by a temperature of the liquid is directly proportional to the difference in temperature of the surrounding.

Multiple choice physics calorimetry heat exchange calorimeter measuring thermal quantities by the method of mixtures

2000 J of energy is needed to heat 1 kg of paraffin through $1^{\circ}C$. So How much energy is needed to heat 10 kg of paraffin through $2^{\circ}C$ ?

  1. 4000 J

  2. 10,000 J

  3. 20,000 J

  4. 40,000 J
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$W= mc\theta$
$2000= (1000)c(1)$
$c= 2$ $J/g^oC$

we get value of c
Hence for $10 kg$ through $2^oC$,
$W= (10000)(2)(2)= 40000 J$

Multiple choice physics measurement and effects of heat heat exchange calorimeter measuring thermal quantities by the method of mixtures

When in thermal contact, the quantity of heat lost by the hotter body is ...... the amount of heat gained by the colder body. (neglect loss of heat due to convection & radiation)

  1. Equal to

  2. Greater than

  3. Less than

  4. Cannot say
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
When in thermal contact, the quantity of heat lost by the hotter body is equal to the amount of heat gained by the colder body. (neglect loss of heat due to convection & radiation)
This is the principle of calorimetry.