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 the essence of change different forms of energy energy conversions energy transformations and energy transfers

A lead ball moving with velocity v strikes a wall and stops. I 50% of its energy is converted into heat, then what will b the increase in temperature? (Specific heat of lead is s)

  1. $\dfrac { 2{ v }^{ 2 } }{ Js } $
  2. $\dfrac { { v }^{ 2 } }{ 4Js } $
  3. $\dfrac { { v }^{ 2 }s }{ J } $
  4. $\dfrac { { v }^{ 2 }s }{ 2J } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

0.5 * (0.5 * m * v^2) = m * s * delta T * J. 0.25 * m * v^2 = m * s * delta T * J. delta T = v^2 / (4 * J * s).

Multiple choice physics electric current and its effects thermal effect of electric current heating effect of electric current effect of electric current

Fill in the blank.

The heat generated in calorie is equal to _________.

  1. $ VIt$
  2. $(VIt)/4.2 $
  3. $(V^2Rt)/4.2 $
  4. $VI^2t$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The right matches are given below.
Heat Generated$\rightarrow $Propotional to the square of current$\rightarrow $ $\frac { VIt }{ 4.18 } cal$  
Resistance in parallel$\rightarrow $Is used to reduce effective resistance in a circuit$\rightarrow $  $\frac { 1 }{ { R } _{ p } } =\frac { 1 }{ { R } _{ 1 } } +\frac { 1 }{ { R } _{ 2 } } $
Resistivity$\rightarrow $Depends on the material of the conductor$\rightarrow $  $\rho =\frac { RA }{ l } $
Ohm's law$\rightarrow $Gives relation between V and I$\rightarrow $V=IR

Multiple choice physics electric current and its effects thermal effect of electric current heating effect of electric current effect of electric current

Rate of dissipation of Joules heat in resistance per unit volume is (symbols have usual meaning)

  1. $\sigma E$
  2. $\sigma J$

  3. J E

  4. None
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The differential form of the Joule heating equation gives the power per unit volume.

$\cfrac {dp}{dv}=J.E$
Here, $J=$ Current Density
          $E=$ Electric Field.

Multiple choice physics electric current and its effects thermal effect of electric current heating effect of electric current effect of electric current

If the current is flowing through a $10 \Omega$ resistor, then in which case the maximum heat will be generated?

  1. $5 \ ampere \ in \ 2 \ minutes$
  2. $4 \ ampere \ in \ 3 \ minutes$
  3. $3 \ ampere \ in \ 6\ minutes$
  4. $2 \ ampere \ in \ 5 \ minutes$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Heat generated $H = i^2 Rt$

(A) : $H _A = (5)^2 (10) (2\times 60) = 30 kJ$
(B) : $H _B = (4)^2 (10) (3\times 60) = 28.8 kJ$

(C) : $H _C = (3)^2 (10) (6\times 60) = 32.4 kJ$
(D) : $H _D = (2)^2 (10) (5\times 60) = 12 kJ$

Thus maximum heat is generated in case $C$.

Multiple choice physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

A $500\ W$ heating unit is designed to operate on a $115\ V$ line. If line voltage drops to $110\ V$ line, the percentage drop in heat output will be:

  1. $7.6\ \%$
  2. $8.5\ \%$
  3. $8.1\ \%$
  4. $10.2\ \%$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given:
$H _1 = 500\ W$
$V _1 = 115\ V$
$V _2 = 110\ V$

From Joule's Law of heating,
$H _1 = \cfrac{(V _1)^2}{R}$ and $H _2 = \cfrac{(V _2)^2}{R}$
$\Rightarrow R = \cfrac{(V _1)^2}{H _1} = \cfrac{(V _2)^2}{H _2}$
$\Rightarrow H _2 = \cfrac{V _2^2}{V _1^2} H _1$
$\therefore H _2 = \cfrac{(110)^2}{(115)^2} (500)$
$\therefore H _2 = 457.46 W$

The percentage drop in heat output will be:
$\cfrac{H _2-H _1}{H _1} \times 100 = \cfrac{500 - 457.46}{500} \times 100 = \cfrac{42.54}{500} \times 100 = 8.5\ \%$

Multiple choice physics electrical circuits domestic electric circuits and safety precautions domestic and commercial circuit electric power

A house hold $20$ liter geyser consist of a coil of resistance $R$, across which an $ac$ source of $rms$ voltage $V$ is connected. The wall thickness of geyser is $t(< < \sqrt{A})$ and the inner surface area of geyser is $A$. The average thermal conductivity of material of walls of geysers is $k$ and $S$ is specific heat capacity of water. Assuming initial temperature of water is equal to the temperature of atmosphere. The time taken to rise the temperature of water by $50^{o}C$ is (Neglect the specific heat of wall of geyser and radiation losses)

  1. $\dfrac{2St}{kA} In \left(\dfrac{50RkA}{rV^{2}}-1\right)$
  2. $\dfrac{20St}{kA} In \left(\dfrac{tV^{2}}{V^{2}t-20RkA}-1\right)$
  3. $\dfrac{20St}{kA} In \left(\dfrac{50tV^{2}}{RkA}\right)$
  4. $\dfrac{20St}{kA} In \left(\dfrac{50RkA}{tV^{2}}\right)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a complex thermal physics problem involving heat input and heat loss through the walls. The expression in option A correctly models the logarithmic time dependence for heating a system with losses.

Multiple choice physics electrical circuits domestic electric circuits and safety precautions domestic and commercial circuit electric power

A bulb made of tungsten filament of surface area $0.5 c{ m }^{ 2 }$ is heated to a temperature 3000 k when operated at 220 V. The emissivity of the filament is $ \in =0.35$ and take $\sigma =5.7\times 1{ 0 }^{ -8 }$ mks units. Then the wattage of the bulb is .. (calculate)

  1. 80.8 W

  2. 0.81 W

  3. 81.2 W

  4. 8.12 W
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The power radiated by a black body is given by the Stefan-Boltzmann law: P = emissivity * sigma * Area * T^4. Plugging in the values: 0.35 * 5.7e-8 * (0.5e-4 m^2) * (3000)^4. Calculating this yields approximately 80.8 W.

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

Photons absorbed in meter are converted to heat. A source emitting $n$ photons/s of frequency $v$ is used to convert $1\ kg$ of ice of ${0}^{o}c$ to water at ${0}^{o}C$. Then, the time taken for the conversion:

  1. decreases with increasing $n$, with $v$ fixed
  2. decreases with $n$ fixed, $v$ increasing
  3. remains constant with $n$ and $v$ changing such that $nv=$constant
  4. increases when the product $nv$ increases
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The energy absorbed per second is P = n * h * v. Since the energy required to melt the ice is constant, the time taken is inversely proportional to the power P. Thus, increasing n (with v fixed) increases power and decreases time.

Multiple choice physics heat energy transfers heat and heat transfer heat energy transfer transfer of heat

Two thin walled sphere of different materials ,one with double the radius and one fourth wall thickness of the other are filled with ice If the time taken for complete melting of ice in the sphere of larger radius is $25$ minutes and that for smaller one is $16$ minutes,the ratio of thermal conductivities of the materials of larger sphere to the smaller sphere is  

  1. $4:5$
  2. $25:1$
  3. $1:25$
  4. $8:25$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the heat conduction formula Q = (KA dT t) / d, where Q is proportional to mass (volume * density), we relate the time taken to melt ice to the thermal conductivity. The ratio of conductivities is derived from the given radii, thicknesses, and times.

Multiple choice physics heat energy transfers heat and heat transfer heat energy transfer transfer of heat

Two rods having thermal conductivities in the ratio of 5:3 and having equal length length and equal cross-section are joined by face to face. If the temperature of free end of first rod is $100^oC$ and temperature of  free end of second rod is $20^oc$, then temperature of the junction, is-

  1. $90^oC$
  2. $85^oC$
  3. $70^oC$
  4. $50^oC$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the thermal conductivity of first rod be $5 K$

Let the thermal conductivity of second rod be $3 k$
now, sum of heat current flowing through the junction =0
$\begin{array}{l} \frac { { 5KA\left( { 100-T } \right)  } }{ x } +\frac { { 3KA\left( { 20-T } \right)  } }{ x } =0 \ 500-5T+60-3T=0 \ 8T=560 \ \therefore T={ 70^{ 0 } }C \end{array}$

Multiple choice physics heat energy transfers heat and heat transfer heat energy transfer transfer of heat

Two walls of thickness $d _1$ and $d _2$, thermal conductivities $K _1$ and $K _2$ are in contact. In the steady state if the temperatures at the outer surfaces are $T _1$ and $T _2$, the temperature at the common wall will be

  1. $\dfrac{K _1T _1+K _2T _2}{d _1+d _2}$
  2. $\dfrac{K _1T _1d _2+K _2T _2d _1}{K _1d _2+K _2d _1}$
  3. $\dfrac{(K _1d _1+K _2d _2)T _1T _2}{T _1+T _2}$
  4. $\dfrac{K _1d _1T _1+K _2d _2T _2}{K _1d _1+K _2d _2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In steady state, the heat flow rate through both layers must be equal. Solving the equation (K1/d1)(T1 - T) = (K2/d2)(T - T2) for T gives the interface temperature.

Multiple choice physics heat energy transfers heat and heat transfer heat energy transfer transfer of heat

Spheres $P$ and $Q$ are uniformly constructed from the same material which is a good conductor of heat and the radius of $Q$is thrice the radius of $P$. The rate of fall of temperature of $P$ is $x$ times that of $Q$ when both are at the same surface temperature. The value of $x$ is:

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

The rate of cooling (dT/dt) is proportional to the surface area divided by the volume (A/V). For a sphere, A/V = (4*pi*r^2) / ((4/3)*pi*r^3) = 3/r. Since the radius of Q is 3 times that of P, the rate of cooling for P is 3 times the rate for Q (x = 3/1 = 3).