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 effects of electric current joule's law of heating thermal effect of electric current heating effect of electric current

What is the Joule's law ?

  1. $H=VRt$
  2. $H=I^2R^2t$
  3. $H=I^2Rt$
  4. $H=\dfrac{V^2}{R}t$
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation

Joules law can be stated as the quantity of heat generated (H) in a conductor of Resistance (R), when a current (I) flows through it for a time (t) is directly proportional to the square of the current, the resistance of the conductor, and the time for which the current flows.
So, $H=I^2Rt$
Now, for a conductor following Ohm's Law, $V=IR$, or, $I=\dfrac{V}{R}$
Therefore, $H=\dfrac{V^2}{R^2} \times R \times t=\dfrac{V^2}{R}t$

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

A given quantity of water boils in an electric kettle in $12 min$. The length of the heating element in the kettle is $\iota $. If the same quantity of water is to boil in $10 min$ on the same mains the length of an identical heating element is _______.

  1. $\iota $
  2. $\displaystyle \frac {\iota }{2}$
  3. $\displaystyle \frac {3\iota }{4}$
  4. $\displaystyle \frac {5\iota }{6}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Amount of heat required to boil a given quantity of water in a constant

$\displaystyle P=\frac {W}{t}$ or $W = H = P \times t = \displaystyle \frac {V^2}{R} \times t = \displaystyle \frac {V^2t _1}{R _1} = \displaystyle \frac {V^2t _2}{R _2}$

$\displaystyle \Rightarrow \frac {t _2}{t _1} = \frac {R _2}{R _1} = \frac {l _2}{l _1}$

or $\displaystyle \frac {10}{12} = \frac {l _2}{l _1} \Rightarrow l _2 = \frac {10}{12} \times l _1$

$\displaystyle l _2 = \frac {5}{6}l$

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

How much heat, in joules, must be added to 0.250 mol of Ar(g) to raise its temperature from 20.0 to $36.0^\circ C$ of at constant pressure?

  1. $50.0J$
  2. $83.14J$
  3. $18J$
  4. $200J$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

At constant pressure,
$q=nCp  \Delta t$               $C _p =\frac{5R}{2}$
$=(0.25) \left ( \frac{5}{2} \times 8.314  \right ) (16)$
$=83.14 J$

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

Heat produced by the resistance R is:

  1. $\displaystyle \frac { VIt }{ 4.2 } $
  2. $\displaystyle \frac { W }{ J } $
  3. $\displaystyle \frac { { i }^{ 2 }Rt }{ 4.2 } $

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

From Joule's Law of heating-


Heat produced$=H=I^2Rt$

Now, $V=IR$

$\implies H=(IR)It=VIt$

And, $I=\dfrac{V}{R}$

$H=I^2Rt=\dfrac{V^2}{R^2}Rt=\dfrac{V^2}{R}t$

Multiple choice physics transfer of heat applications of convection convection thermal energy transfers

A double-plane window consist of two glass sheets each of area $1m^2$ and thickness $0.01m$ separated by a $0.05m$ thick stagnant air space In the steady state, the room glass interface and the glass outdoor interface are at constant temperature of $27^oC$ and $0^oC$ respectively. The rate of heat flow through the window plane is (Given , $k _{glass}=0.8\,\,W\,\,m^{-1}K^{-1},K _{air}=0.08\,\,W\,\,m^{-1}K^{-1})$

  1. $41.5\,\,W$
  2. $31.5\,\,W$
  3. $21.5\,\,W$
  4. $11.5\,\,W$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total thermal resistance is 
$R=\dfrac{t _1}{K _1A _1}+\dfrac{t _2}{K _2A _2}+\dfrac{t _1}{K _1A _1}$


$R=2\times \dfrac{0.01}{0.8\times 1}+\dfrac{0.05}{0.08\times 1}= 0.65W^{-1}K$

$\therefore $Heat current $,H=\dfrac{\triangle T}{R}=\dfrac{27-0}{0.65}=41.5W$

Multiple choice physics transfer of heat applications of convection convection thermal energy transfers

Heat is flowing through two cylindrical rods made of same materials whose ends are maintained at similar temperatures. If diameters of the rods are in ratio 1 : 2 and lengths in the ratio 2 : 1, then the ratio of thermal current through in steady state is:

  1. 1 : 8

  2. 1 : 4

  3. 1 : 6

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

The thermal current is nothing but the rate of flow heat,


$I=\dfrac{Q}{t}=K\Delta T\dfrac{A}{l}$. . . . . .(1)

Where,

$\Delta T= $ change in temperature

$A=\pi r^2$, cross-sectional area of the cylindrical rod.

$K=$ Thermal conductivity.

$l=$ length of the rod

From equation (1)

Thermal current is $I=4\pi K\Delta T \dfrac{d^2}{l}$

$I\alpha \dfrac{d^2}{l}$ where, $d=$ diameter of the cylindrical rod

Given,

$l _1:l _2=2:1$

$d _1:d _2=1:2$

The ratio of the thermal current  is,

$\dfrac{I _1}{I _2}=(\dfrac{d _1}{d _2})^2\times \dfrac{l _2}{l _1}$

$\dfrac{I _1}{I _2}=(\dfrac{1}{2})^2\times \dfrac{1}{2}$

$\dfrac{I _1}{I _2}=\dfrac{1}{8}$

$I _1:I _2=1:8$

Option A is correct.

Multiple choice physics transfer of heat applications of convection convection thermal energy transfers

Temperature of a room is ${ -15 }^{ 0 }C$ and outside temperature is ${ 10 }^{ 0 }C.$ If room temperature is made ${ 40 }^{ 0 }C$, then find outside temperature, if rate of heat flow is same in both cases. 

  1. ${ 10 }^{ 0 }C$
  2. ${ 5 }^{ 0 }C$
  3. ${ 15 }^{ 0 }C$
  4. ${ 20 }^{ 0 }C$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Rate of heat flow dQ/dt = k(T_hot - T_cold). Case 1: k(10 - (-15)) = k(25). Case 2: k(40 - T_outside) = k(25). Thus, 40 - T_outside = 25, so T_outside = 15 degrees Celsius.

Multiple choice physics transfer of heat applications of convection convection thermal energy transfers

A metallic sphere cools from $50^{\circ}C$ to $40^{\circ}C$ in $300\ s$. If the room temperature is $20^{\circ}C$, then its temperature in the next $5$ min will be

  1. $38^{\circ}C$
  2. $33.3^{\circ}C$
  3. $30^{\circ}C$
  4. $36^{\circ}C$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using Newton's Law of Cooling: (T1 - T2)/t = k(T_avg - T_env). (50 - 40)/300 = k(45 - 20) => 10/300 = k(25) => k = 1/750. For the next 300s: (40 - T)/300 = (1/750)( (40+T)/2 - 20). Solving gives T = 30 degrees Celsius.

Multiple choice physics transfer of heat applications of convection convection thermal energy transfers

A sphere of density d, statisfied heat s and radius r is hung by thermally insulating thread in an enclosure which is kept at a lower temperature than the sphere.The temperature of the sphere to drop at a rate which depends upon the temperature difference between the and the enclosure,If the temperature difference is $ \triangle T $ and surrounding temperature is $ T _o $ then rate of fall in temperature will be-

  1. $ \frac {4 \sigma T _o^2 \triangle T}{rdc} $
  2. $ \frac {12 \sigma T _o^2 \triangle T}{rdc} $
  3. $ \frac {12 \sigma T _o^4 \triangle T}{rdc} $
  4. $ \frac {12\sigma \triangle T}{rdc T _o^3 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice kirchoff's laws black body radiation heat transfer thermal properties physics

If the emissive and absorptive powers of a body are $E$ and $A $ respectively at temperature $T$ then emissive power of a black body will be

  1. $E/A$
  2. $EAT$
  3. $EA/T$
  4. $A/E$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If Emissive power =$ E$
Absorptive power = $A$
The emissive power of a black body is the emissive power for unit absorptive power(per unit absorptive power) (Black body absorbs all radiations)  = $\dfrac{E}{A}$

Multiple choice kirchoff's laws black body radiation heat transfer thermal properties physics

If $p$ calories of heat energy is incident on a body and $q$ calories is absorbed, then its coefficient of absorption is :

  1. $\dfrac{p}{q}$
  2. $p - q$
  3. $\dfrac{q}{p}$
  4. $q + p$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Calories of incident heat=$p$
Calories of absorbed heat=$q$
$\because$ we know that coefficient of absorption is the ratio of heat absorbed to the incident heat.
So, coefficient of absorption=$\cfrac{q}{p}$
Multiple choice kirchoff's laws black body radiation heat transfer thermal properties physics

Given that $p$ Joules of heat is incident on a body and out of it $q$ Joules is reflected and transmitted by it, the absorption coefficient of the body is

  1. $(q-p)/p$
  2. $q/p$
  3. $(p-q)/p$
  4. $p/q$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Coefficient of absorption is defined as the ration of amount of radiation a body absorbs to the amount of radiation that is incident on the body.
Incident radiation = $p$
Rejected radiation = $q$
Hence absorbed radiation must be = $p - q$
From the definition of absorptive power it follows that the power = $\dfrac{p-q}{p}$

Multiple choice kirchoff's laws black body radiation heat transfer thermal properties physics

If the amount of heat incident upon a body is $X$ calorie and it absorbs $Y$ calorie out of it , then the coefficient of absorption will be

  1. $X+Y$
  2. $XY$
  3. $Y/X$
  4. $X/Y$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Coefficient of absorption is defined as the ration of amount of radiation a body absorbs to the amount of radiation that is incident on the body.
Incident radiation = $X$
Absorbed radiation  = $Y$
Hence absorptive power = $\dfrac{Y}{X}$

Multiple choice kirchoff's laws black body radiation heat transfer thermal properties physics

If temperature of a black body increases from $-73^oC$ to $327^oC$, then ratio of emissive power at the two temperature is

  1. 27 : 1

  2. 81 : 1

  3. 1 : 27

  4. 1 : 81
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given,

$T _1=-73^0C=273^0C-73^0C=200K$
$T _2=327^0C=273^0C+327^0C=600K$
According to Stephen's law of radiation,
$E\propto T^4$
Ratio, $\dfrac{E _1}{E _2}=(\dfrac{T _1^4}{T _2^4})^4=(\dfrac{200}{600})^4$
$\dfrac{E _1}{E _2}=\dfrac{1}{81}$
$E _1:E _2=1:81$
The correct option is D.