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 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.

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

The $1847$ Joule's experiment was aimed at.

  1. Determining the mechanical equivalent of heat

  2. Determining the temperature for the maximum density of water

  3. Investigating the heating effect of electric current

  4. Investigating the internal energy of a gas
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

James Prescott Joule performed experiments in the 1840s to demonstrate that heat is a form of energy. His work established the mechanical equivalent of heat, showing that a specific amount of mechanical work could produce a specific amount of heat.

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)  = $\dfrac{E}{A}$