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

If the amount of heat given to a system is $35\, J$ and the amount of work done on the system is $15\, J$, then the change in internal energy of the system is

  1. $- 50\, J$
  2. $20\, J$
  3. $30\, J$
  4. $50\, J$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given,

$\Delta Q=+35J$
$\Delta W=-15J$
$\Delta U=?$
From law of thermodynamic,
$\Delta Q=\Delta U+\Delta W$
$\Delta U=\Delta Q-\Delta W$
$\Delta U=35-(-15)$
$\Delta U=35+15$
$\Delta U=50J$
The correct option is D. 

Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

A geyser heats water flowing at the rate of 3.0 liters per minute from ${ 27 }^{ \circ  }C$ to ${ 77 }^{ \circ  }C$. If the geyser operates on a gas burner, the rate of consumption of the fuel if its heat of combustion is $4.0\times { 10 }^{ 4 }J/g$ per minute is

  1. $15.75g$
  2. $4 g$
  3. $0.3 g$
  4. $0.16 g$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Heat required = mass * specific heat * delta T = 3000 g * 1 cal/g C * 50 C = 150,000 cal = 630,000 J. Fuel consumption rate = Total heat / Heat of combustion = 630,000 J / 40,000 J/g = 15.75 g.

Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

A hammer of mass 1$\mathrm { kg }$ having speed of 50$\mathrm { m } / \mathrm { s }$ , hit a iron nail of mass 200$\mathrm { gm }$ . If specific heat of iron is 0.105 cal/gm'C and half the energy is converted into heat, the raise in temperature of nail is

  1. $7.1 ^ { \circ } C$
  2. $9.2 ^ { \circ } \mathrm { C }$
  3. $10.5 ^ { \circ } C$
  4. $12.1 ^ { \circ } \mathrm { C }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Initial kinetic energy = 0.5 * m * v^2 = 0.5 * 1 * 50^2 = 1250 J. Half of this is 625 J. Convert to calories: 625 / 4.2 = 148.8 cal. Heat = ms(delta T) -> 148.8 = 200 g * 0.105 cal/g C * delta T. delta T = 148.8 / 21 = 7.08 C.

Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

Work done by 100 calorie of heat is __________.

  1. 418.4 J

  2. 4.184 J

  3. 41.84 J

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
For an isothermal expansion of an ideal gas, the change in internal energy is zero.

According to the first law of thermodynamics, 

Change in internal energy U = Q-W = 0

So, all the heat energy is utilized to do work. 

Q = W

We know that, one calorie is equal to 4.184 J

Therefore, Work done by 100 calorie of heat in an isothermal expansion of any ideal gas will be 4.184 * 100 =  418.4 Joule


Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

A block of ice at 0 C whose mass is initially 50.0 kg slides along a horizontal surface starting at a speed of 5.38 m/s and finally coming of ice melted as a result of the friction between the block and the surface will be

  1. 2.16 g

  2. 4.0 g

  3. 1 g

  4. 50 g
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Initial KE = 0.5 * 50 * 5.38^2 = 723.6 J. Heat = 723.6 J. Mass melted = Heat / Latent heat = 723.6 / 334,000 J/kg = 0.00216 kg = 2.16 g.

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

The maximum amount of work that can be done by extracting $1J$ of energy from a body at $127^oC$ with an environment at $27^oC$?

  1. $\dfrac{1}{8}J$
  2. $\dfrac{1}{4}J$
  3. $\dfrac{1}{2}J$
  4. $\dfrac{3}{4}J$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The maximum work obtainable from a heat engine operating between two temperatures is given by Carnot efficiency: eta = 1 - (T_L / T_H). Here T_H = 127 + 273 = 400 K and T_L = 27 + 273 = 300 K. Thus eta = 1 - (300 / 400) = 1/4 = 0.25. For 1 J of energy extracted from the hot body, the maximum work done is W = eta * Q_H = (1/4) * 1 = 1/4 J.

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.