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 temperature and heat modes of heat transfer - conduction conduction heat and modes of heat transfer

A spherical black body of radius $(R)$ when heated to certain temperature and left in vaccum. cools at a rate $'x' $ Now a caity of radius $(R/2)$ is made concentrically from this sphere. The rate of cooling of the remaining sphere will be.....................

  1. $\cfrac { x }{ 8 } $
  2. $\cfrac { 7x }{ 8 } $
  3. $\cfrac { 8x }{ 8 } $
  4. $\cfrac { x }{ 7 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics temperature and heat modes of heat transfer - conduction conduction heat and modes of heat transfer

A hollow copper sphere and a hollow copper cube, of same surface area and negligible thickness, are filled with warm water of same temperature and placed in an enclosure of constant temperature, a few degrees below that of the bodies. Then in the beginning:

  1. the rate of energy lost by the sphere is greater than that by the cube

  2. the rate of fall of temperature for sphere is greater than that for the cube.

  3. the rate of energy lost by the sphere is less than that by the cube

  4. the rate of fall of temperature for sphere is less than that for the cube.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Rate of cooling dT/dt = (sigma * A * (T^4 - Ta^4)) / (m * c). For the same surface area A, the body with the smaller mass m will have a higher rate of temperature fall. A hollow sphere has less volume (and thus less mass) than a hollow cube of the same surface area.

Multiple choice physics temperature and heat modes of heat transfer - conduction conduction heat and modes of heat transfer

A copper block of mass $500gm$ and $Sp.$ Heat $0.1 cal/gm/^{o}{C}$ is heated from ${30}^{o}C$ to ${40}^{o}C$. Another identical copper block $B$ of same mass is heated from ${35}^{o}C$ to ${40}^{o}C$. The ratio of their thermal capacities is 

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

Thermal capacity is defined as mass * specific heat (m * c). Since both blocks have the same mass and are made of the same material (copper), their thermal capacities are identical, resulting in a 1:1 ratio.

Multiple choice physics temperature and heat modes of heat transfer - conduction conduction heat and modes of heat transfer

Three copper blocks of masses ${ M } _{ 1 },{ M } _{ 2 }$ and ${ M } _{ 3 }$ kg respectively are brought into thermal contact till they reach equilibrium. Before contact. they were at ${ T } _{ 1 },{ T } _{ 2 },{ T } _{ 3 }$ $\left( { T } _{ 1 }>{ T } _{ 2 }>{ T } _{ 3 } \right) .$ Assuming there is no heat loss to the surrounding, the equilibrium temperature T (s is specitc heat of copper)

  1. $T=\dfrac { { T } _{ 1 }+{ T } _{ 2 }+{ T } _{ 3 } }{ 3 } $
  2. $T=\dfrac { { M } _{ 1 }{ T } _{ 1 }+{ M } _{ 2 }{ T } _{ 2 }+{ M } _{ 3 }{ T } _{ 3 } }{ { M } _{ 1 }+{ M } _{ 2 }+{ M } _{ 3 } } $
  3. $T=\dfrac { { M } _{ 1 }{ T } _{ 1 }+{ M } _{ 2 }{ T } _{ 2 }+{ M } _{ 3 }{ T } _{ 3 } }{ 3{ (M } _{ 1 }+{ M } _{ 2 }+{ M } _{ 3 }) } $
  4. $T=\dfrac { { M } _{ 1 }{ T } _{ 1 }s+{ M } _{ 2 }{ T } _{ 2 }s+{ M } _{ 3 }{ T } _{ 3 }s }{ { M } _{ 1 }+{ M } _{ 2 }+{ M } _{ 3 } } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics temperature and heat modes of heat transfer - conduction conduction heat and modes of heat transfer

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. $\dfrac { K _ { 1 } T _ { 1 } + K _ { 2 } T _ { 2 } } { d _ { 1 } + d _ { 2 } }$
  2. $\dfrac { K _{ { 1 } }T _{ 1 }d _{ { 2 } }+K _{ { 2 } }T _{ { 2 } }d _{ { 1 } } }{ K _{ { 1 } }d _{ { 2 } }+K _{ { 2 } }d _{ { 1 } } } $
  3. $\dfrac { \left( K _ { 1 } d _ { 1 } + K _ { 2 } d _ { 2 } \right) T _ { 1 } T _ { 2 } } { T _ { 1 } + T _ { 2 } }$
  4. $\dfrac { K _{ { 1 } }d _{ { 1 } }T _{ 1 }+K _{ { 2 } }d _{ { 2 } }T _{ { 2 } } }{ K _{ { 1 } }d _{ { 1 } }+K _{ { 2 } }d _{ { 2 } } } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{array}{l} Under\, steady\, state\, heat\, flux\, per\, unit\, area\, k\left( { \frac { { dT } }{ { dx } }  } \right)  \ is\, same\, across\, two\, walls.\, hence,\, we\, have \ { K _{ 1 } }\dfrac { { { T _{ 1 } }-{ T _{ c } } } }{ { { d _{ 1 } } } } ={ K _{ 2 } }\dfrac { { { T _{ c } }-{ T _{ 2 } } } }{ { { d _{ 2 } } } }  \ where\, { T _{ c } }\, is\, common\, wall\, temperature.\, \, solving\, for\, { T _{ c } }\, we\, will\, get \ { T _{ c } }=\dfrac { { { T _{ 1 } }+\alpha { T _{ 2 } } } }{ { \alpha +1 } }  \ Where\, \alpha =\dfrac { { { d _{ 1 } } } }{ { { d _{ 2 } } } } \, \dfrac { { { k _{ 2 } } } }{ { { k _{ 2 } } } }  \end{array}$

$\begin{array}{l} On\, putting\, the\, value\, of\, \alpha =\dfrac { { { d _{ 1 } } } }{ { { d _{ 2 } } } } .\dfrac { { { k _{ 1 } } } }{ { { k _{ 2 } } } }  \ Then,\, { T _{ c } }=\dfrac { { { k _{ 1 } }{ T _{ 1 } }{ d _{ 2 } }+{ k _{ 2 } }{ T _{ 2 } }{ d _{ 1 } } } }{ { { k _{ 1 } }{ d _{ 2 } }+{ k _{ 2 } }{ d _{ 1 } } } }  \end{array}$
Hence,Option $B$ is the correct answer.

Multiple choice physics temperature and heat modes of heat transfer - conduction conduction heat and modes of heat transfer

Two rods of length  $\mathrm { d _ { 1 } } ,$  and  $\mathrm { d _ { 2 } } ,$  and coefficient of thermal conductivities  $\mathrm { K } _ { 1 }$  and  $\mathrm { K } _ { 2 }$  are kept touching each other. Both have the same area of cross-section. The equivalent of thermal conductivity is

  1. $K _ { 1 } + K _ { 2 }$
  2. $\mathrm { K } _ { 1 } \mathrm { d } _ { 1 } + \mathrm { K } _ { 2 } \mathrm { d } _ { 2 }$
  3. $\dfrac { \mathrm { d } _ { 1 } \mathrm { K } _ { 2 } + \mathrm { d } _ { 2 } \mathrm { K } _ { 2 } } { \mathrm { d } _ { 1 } + \mathrm { d } _ { 2 } }$
  4. $\dfrac { d _ { 1 } + d _ { 2 } } { \left( d _ { 1 } K _ { 2 } \right) + \left( d _ { 2 } K _ { 2 } \right) }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice physics temperature and heat modes of heat transfer - conduction conduction heat and modes of heat transfer

Three roads identical area of cross-section and made from the same metal from the sides of an isosceles triangle ABC, right angled at B. The points A and B are maintained at temperature  T and $ \sqrt {2} T $ respectively. IN the steady state the temperature that only point C is $ T _c $ Assuming that only conduction takes place $ \frac {T _c}{T} is $

  1. $ \frac { 1 }{ \left( \sqrt { 2 } +1 \right) } $
  2. $ \frac { 1 }{ \left( \sqrt { 2 } -1 \right) } $
  3. $ \frac { 1 }{ 2\left( \sqrt { 2 } +1 \right) } $
  4. $ \frac { 1 }{ \sqrt { 3 } \left( \sqrt { 2 } -1 \right) } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics temperature and heat modes of heat transfer - conduction conduction heat and modes of heat transfer

Two rods of equal length and area of cross-sectional are kept parallel and lagged between temperature $ 20^o C and 80^oC $ The ration of the effective thermal conductivity to that of the first rod is
 $ \left[ the\quad ration\left( \frac { K _ 1 }{ K _ 2 }  \right) =\frac { 3 }{ 4 }  \right]  $

  1. 7 : 4

  2. 7 : 6

  3. 4 : 7

  4. 7 : 8
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics temperature and heat modes of heat transfer - conduction conduction heat and modes of heat transfer

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

  1. $4:5$
  2. $5:4$
  3. $25:1$
  4. $1:25$
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice physics temperature and heat modes of heat transfer - conduction conduction heat and modes of heat transfer

If a rod is in a variable state (not in steady state), then

  1. Temperature gradient remains constant

  2. Temperature of rod is function of time and distance from one of end

  3. Temperature of rod is only function of distance from one of end

  4. Temperature of rod is only function of time
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In a variable state (non-steady state), the temperature at any point in the rod changes with time as heat is absorbed by the material to change its internal energy. Therefore, temperature is a function of both position and time.

Multiple choice physics temperature and heat modes of heat transfer - conduction conduction heat and modes of heat transfer

The quantity of heat flowing for $10 \ s$ through a rod of length $40\ cm$, area $50 \ cm^{2}$ is $200\ J$. If the temperature difference at the ends of the rod is 80$^{o}$C , the coefficient of thermal conductivity of the rod in Wm$^{-1}$K$^{-1}$ is:

  1. $20$
  2. $60$
  3. $80$
  4. $120$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$Q = 200$
$A = 50 \times 10^{-4}\ m^2$
${\theta}^{} _{1} = 80^\circ C    ;   \ \ {\theta}^{} _{2} = 0$
$t = 10\ s ; \ \    l  = 0.40\ m$
$Q = \dfrac{KA\Delta T t}{L}$
$200 = \dfrac{K  \times 50\times 10^{-4}\times    80\times    10}{0.4}$ 
$K = 20\ W/mK$ 

Multiple choice physics temperature and heat modes of heat transfer - conduction conduction heat and modes of heat transfer

A piece of metal is heated to increase its temperature from $5^{\circ}C$ to $15^{\circ}C$. The increase in temperature expressed in $K$ and $^{\circ}F$ are respectively.

  1. $10\ K, 18^{\circ}F$
  2. $283\ K, 50^{\circ}F$
  3. $18\ K, 10^{\circ}F$
  4. $50\ K, 283^{\circ}F$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$K=273.16+C\F=\cfrac{9C}{5}+32\quad dK=dC \quad dF=\cfrac{9}{5}dC$

A/Q, $dC=15-5=10\ \therefore dK=10K,\quad dF=\cfrac{9}{5}\times10=18°F$

Multiple choice physics temperature and heat modes of heat transfer - conduction conduction heat and modes of heat transfer

A slab of stone area $3500{cm}^{2}$ and thickness $10cm$ is exposed on the lower surface to steam at ${100}^{o}C$. A block of ice at ${0}^{o}C$ rests on upper surface of the slab. In one hour $4.8kg$ of ice of melted. The thermal conductivity of the stone is $J{s}^{-1}$ ${m}^{-1}$ ${k} _{-1}$ is
(Latent heat of ice $=3.36\times { 10 }^{5 }J/kg$)

  1. $12.0$
  2. $10.5$
  3. $1.02$
  4. $1.24$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given :  $A = 3500 \ cm^2 = 0.35 \ m^2$         $l = 10 \ cm = 0.1 \ m$             $\Delta T =100-0 = 100^o C$
Mass of ice melted  $m = 4.8 \ kg$
Time taken  $t = 1 \ hr = 3600 \ s$
Latent heat of ice  $L = 3.36\times 10^{5} \ J/kg$
Heat absorbed by ice = Heat conducted by slab
$\therefore$   $mL = \dfrac{KA t\Delta T}{l}$
Or    $4.8\times 3.36\times 10^{5} = \dfrac{K(0.35) (3600)(100)}{0.1}$
$\implies \ K = 1.24 \ Js^{-1} m^{-2} k^{-1}$