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 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).

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

A metal cylinder of mass 0.5 kg is heated electrically by a 12 W heater in a room and cylinder temperature rises uniformly to $ 25^oC in 5 min $ excess temperature surroundings,

  1. The rate of loss of heat of the cylinder to surrounding at $ 20^oC $ is 2 W
  2. The rate of loss heat of the cylinder to surrounding at $ 45^oC $ is 12 W
  3. Specific heat capacity of metal is $ \frac {240}{Ib(3/2)} $

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice physics heat energy transfers heat and heat transfer heat energy transfer transfer of heat

In the Ingen Hausz's experiment, the wax melts up to lengths $10cm$ and $25cm$ on two identical rods of different materials. The ratio of thermal conductivities of the two materials is

  1. $1:6.25$
  2. $6.25:1$
  3. $1: $$\sqrt{2.5}$
  4. $1:2.5$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In the Ingen Hauszs experiment, as rate of heat transfer is constant
$\dfrac { { K } _{ 1 } }{ { K } _{ 2 } } =\dfrac { { l } _{ 1 } }{ { { l } _{ 2 } } } $
So, putting the value in above formula's we find
$\dfrac{{K} _{1}}{{K} _{2}}=\dfrac{10}{25}=\dfrac{1}{2.5}$

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

Two blocks of steel A and B, A being two times heavier than B, are at 40$^o$C. The ratio of heat content of A to B is:

  1. 1

  2. 4

  3. 2
  4. $\displaystyle \frac{1}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let the mass of block B be $m$.
So, mass of block A is $2m$
Both are made of steel, so, both has same specific heat capacity (let 'S')

So, $\cfrac{Heat\;Capacity\;(A)}{Heat\;Capacity\;(B)}=\cfrac{2ms}{1ms}=2$
Multiple choice chemistry chemical thermodynamics system and surroundings introduction to thermodynamics basics of thermodynamics

Temperatures of two hot bodies $B _{1}$ and $B _{2}$ are $100^{\circ}C$ and $80^{\circ}C$ respectively. The temperature of surrounding is $40^{\circ}C$. At $t = 0$, the ratio of rates of cooling of the two bodies (liquid) $R _{1} : R _{2}$ will be:

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

Rate of cooling=$\cfrac { dQ }{ dt } \propto \quad \Delta T$ 

$\therefore \cfrac { { R } _{ 1 } }{ { R } _{ 2 } } =\cfrac { { \Delta T } _{ 1 } }{ { \Delta T } _{ 2 } } =\cfrac { 100-40 }{ 80-40 } =\cfrac { 60 }{ 40 } =\cfrac { 3 }{ 2 } $

Multiple choice chemistry chemical thermodynamics system and surroundings introduction to thermodynamics basics of thermodynamics

Which of the following statements are not true?

  1. Heat is a macroscopic physical property

  2. Heat is an intrinsic property of a body

  3. Heat is stored in a body as internal energy

  4. Heat is path independent
Reveal answer Fill a bubble to check yourself
B,D Correct answer
Explanation

An intrinsic property is a property of a system or of a material itself or within. It is independent of how much of the material is present and is independent of the form of the material. Heat flow is a results of a temperature difference between two bodies hence, it is not an intrinsic property of substance.

Two important examples of a path function are heat and work. These two functions are dependent on how the thermodynamic system changes from the initial state to final state.

Multiple choice chemistry chemical thermodynamics system and surroundings introduction to thermodynamics basics of thermodynamics

The work done in an open vessel at $300$K, when $112g$ iron reacts with dilute $HCl$ to give $FeCl _2$, is nearly:

  1. $1.1$ kcal
  2. $0.6$ kcal
  3. $0.3$ kcal
  4. $0.2$ kcal
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The reaction involved is:
$Fe+2HCl \rightarrow FeCl _2+H _2$
Atomic mass of $Fe=56 g/mol $
Thus, 56 g of Iron reacts with 2 moles of HCl to give one mole of Hydrogen gas.
Initial volume of $H _2$ gas = $V _1 = 0$
Final volume of $H _2$ gas=$ V _2$
Using ideal gas law:$PV = n R T$
where n=mass/molar mass, R=8314 J/K/mol and given that T=300 K
$PV _2 = (112/56) \times 8.314 \times 300 = 4988.4J$
Work done $= -P \Delta V=-P(V _2 - V _1) =-P V _2 = -4988.4J$
negative work done is work of expansion.
since 4184 J=1 kcal
thus $4988.4 J=1.19 kcal$ of work is done by the system.
Multiple choice chemistry chemical thermodynamics system and surroundings introduction to thermodynamics basics of thermodynamics

An open vessel containing air is heated from 300 K to 400 K. The fraction of air originally present which goes out of it is:

  1. 3/4

  2. 1/4

  3. 2/3

  4. 1/8
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
 
According to ideal gas equation
$PV = nRT$
$\Rightarrow \; n \propto \cfrac{1}{T}$
$\Rightarrow \; \cfrac{{n} _{1}}{{n} _{2}} = \cfrac{{T} _{2}}{{T} _{1}}$
Given that:-
${n} _{1} = 1$ 
${T} _{1} = 300K$, 
${T} _{2} = 400K$ 
${n} _{2} = ?$
$\therefore \; \cfrac{1}{{n} _{2}} = \cfrac{400}{300}$
${n} _{2} = \cfrac{3}{4}$
The fraction of air present in the vessel after heating ${n} _{2} = \cfrac{3}{4}$
The fraction of air which goes out of the vessel $= 1 - \cfrac{3}{4} = \cfrac{1}{4}$