Q 1 :

A water cooler of storage capacity 120 litres can cool water at a constant rate of P watts. In a closed circulation system (as shown schematically in the figure), the water from the cooler is used to cool an external device that generates constantly 3 kW of heat (thermal load). The temperature of water fed into the device cannot exceed 30°C and the entire stored 120 litres of water is initially cooled to 10°C. The entire system is thermally insulated. The minimum value of P (in watts) for which the device can be operated for 3 hours is                                      [2016]

[IMAGE 410]

(Specific heat of water is 4.2 kJ kg-1K-1 and the density of water is 1000 kg m-3)

  • 1600

     

  • 2067

     

  • 2533

     

  • 3933

     

(2)

Pheater-Pcooler=mcΔTt=VρcΔTt

  (3000-Pcooler)=0.12×1000×4.2×103×203×60×60

  Pcooler=2067 W



Q 2 :

Parallel rays of light of intensity I=912 Wm-2 are incident on a spherical black body kept in surroundings of temperature 300 K. Take Stefan-Boltzmann constant σ=5.7×10-8 Wm-2K-4 and assume that the energy exchange with the surroundings is only through radiation. The final steady state temperature of the black body is close to                     [2014]

  • 330 K

     

  • 660 K

     

  • 990 K

     

  • 1550 K

     

(1)

Let T be the final steady state temperature of the black body.

In steady state,

Energy lost = Energy gained

σ(T4-T04)×4πR2=I(πR2)

  5.7×10-8[T4-(300)4]×4=912

  T=330 K



Q 3 :

Three rods of Copper, Brass and Steel are welded together to form a Y shaped structure. Area of cross-section of each rod = 4 cm2. End of copper rod is maintained at 100°C whereas ends of brass and steel are kept at 0°C. Lengths of the copper, brass and steel rods are 46, 13 and 12 cm respectively. The rods are thermally insulated from surroundings except at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 CGS units respectively. Rate of heat flow through copper rod is:       [2014]

  • 1.2 cal/s

     

  • 2.4 cal/s

     

  • 4.8 cal/s

     

  • 6.0 cal/s

     

(3)

[IMAGE 411]

Rate of heat flow is given by,

Q=KA(θ1-θ2)l

Where, K = coefficient of thermal conductivity

l = length of rod  and  A = area of cross-section of rod

If the junction temperature is T, then

QCopper=QBrass+QSteel

0.92×4(100-T)46=0.26×4(T-0)13+0.12×4(T-0)12

  200-2T=2T+TT=40°C

  QCopper=0.92×4×6046=4.8 cal/s



Q 4 :

Two rectangular blocks, having identical dimensions, can be arranged either in configuration-I or in configuration-II as shown in the figure. One of the blocks has thermal conductivity k and the other 2k. The temperature difference between the ends along the x-axis is the same in both the configurations. It takes 9 s to transport a certain amount of heat from the hot end to the cold end in the configuration-I. The time to transport the same amount of heat in the configuration-II is             [2013]

[IMAGE 412]

  • 2.0 s

     

  • 4.5 s

     

  • 3.0 s

     

  • 6.0 s

     

(1)

Equivalent thermal resistance in configuration-I

Rs=R1+R2=LKA+L2KA=32LKA

Equivalent thermal resistance in configuration-II

1Rp=1R1+1R2=KAL+2KAL   or,    Rp=L3KA=Rs4.5

i.e., thermal resistance in configuration-II, Rp is 4.5 times less than thermal resistance in configuration-I, Rs.

  4.5tp=ts  tp=ts4.5=94.5s=2 s



Q 5 :

Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is T, density of liquid is ρ and L is its latent heat of vaporization.          [2012]

  • ρLT

     

  • TρL

     

  • TρL

     

  • 2TρL

     

(4)

When radius is decreased by ΔR,

Decrease in surface energy = Heat lost as latent heat

(4πR2ΔRρ)L=4π[R2-(R-ΔR)2]T

  ρR2ΔRL=T[R2-R2+2RΔR-ΔR2]

  ρR2ΔRL=T(2RΔR)    [ΔR is very small]

  R=2TρL



Q 6 :

Water of volume 2 litre in a container is heated with a coil of 1 kW at 27°C. The lid of the container is open and energy dissipates at rate of 160 J/s. In how much time temperature will rise from 27°C to 77°C [Given specific heat of water is 4.2kJ/kg]                [2005]

  • 7 min

     

  • 6 min 2s

     

  • 8 min 20s

     

  • 14 min

     

(3)

[IMAGE 413]

As shown in the figure, the net heat gained by the water to raise its temperature

=(1000-160)=840 J/s

Now, the heat required to raise the temperature of water from 27°C to 77°C

Q=mcΔt=2×4200×50 J

Hence the time required to gain Q amount of heat

t=Q840=2×4200×50840=500 s=8 min 20 s



Q 7 :

Calorie is defined as the amount of heat required to raise temperature of 1 g of water by 1°C and it is defined under which of the following conditions?          [2005]

  • From 14.5°C to 15.5°C at 760 mm of Hg

     

  • From 98.5°C to 99.5°C at 760 mm of Hg

     

  • From 13.5°C to 14.5°C at 76 mm of Hg

     

  • From 3.5°C to 4.5°C at 76 mm of Hg

     

(1)

1 Calorie is the amount of heat required to raise temperature of 1 gram of water from 14.5°C to 15.5°C at 760 mm of Hg.



Q 8 :

In which of the following process, convection does not take place primarily               [2005]

  • sea and land breeze

     

  • boiling of water

     

  • heating air around a furnace

     

  • warming of glass of bulb due to filament

     

(4)

Warming of glass of bulb due to filament is primarily due to radiation. A medium is required for convection process. As a bulb is almost evacuated, heat from the filament is transmitted through radiation.



Q 9 :

Variation of radiant energy emitted by sun, filament of tungsten lamp and welding arc as a function of its wavelength is shown in figure. Which of the following option is the correct match ?                              [2005]

[IMAGE 414]

  • Sun-T3, tungsten filament-T1, welding arc-T2

     

  • Sun-T2, tungsten filament-T1, welding arc-T3

     

  • Sun-T3, tungsten filament-T2, welding arc-T1

     

  • Sun-T1, tungsten filament-T2, welding arc-T3

     

(1)

[IMAGE 415]

According to Wein's displacement law

λm×T=constant

From the graph, λm3<λm2<λm1

  T3>T2>T1

The temperature of Sun -T3 is higher than that of welding arc T2 which in turn is greater than tungsten filament -T1.



Q 10 :

Two identical rods are connected between two containers, one of them is at 100°C and another is at 0°C. If rods are connected in parallel then the rate of melting of ice is q1 gm/sec. If they are connected in series then the rate is q2. The ratio q2q1 is               [2004]

  • 2

     

  • 4

     

  • 1/2

     

  • 1/4

     

(4)

In parallel combination of rods

KP=K1+K2=K+K=2K

[IMAGE 416]

In series combination

KS=K1-K2K1+K2=KK2K=K2

[IMAGE 417]

q1=2KA(100)  and  q2=KA(100)2

  q2q1=KA(100)2×2KA(100)=14



Q 11 :

If liquefied oxygen at 1 atmospheric pressure is heated from 50 K to 300 K by supplying heat at constant rate. The graph of temperature vs time will be     [2004]

  • [IMAGE 418]

     

  • [IMAGE 419]

     

  • [IMAGE 420]

     

  • [IMAGE 421]

     

(3)

From 50 K to boiling temperature, T increases linearly as Q=mcΔT.  Hence T-t graph will be a straight line inclined to time axis.

During boiling, Q=mL

Temperature remains constant till boiling is complete and graph will be a straight line parallel to time axis.

After that, temperature increases linearly. T-t graph will be a straight line inclined to time axis.



Q 12 :

Three discs A, B and C having radii 2, 4, and 6 cm respectively are coated with carbon black. Wavelength for maximum intensity for the three discs are 300, 400 and 500 nm respectively. If QA, QB and QC are power emitted by A, B and C respectively, then                       [2004]

  • QA will be maximum

     

  • QB will be maximum

     

  • QC will be maximum

     

  • QA=QB=QC

     

(2)

From Wein's displacement law

λmT=Constant

  T=constantλm                 { TA=C3×10-7, TB=C4×10-7, TC=C5×10-7}

Again from Stefan's law

Q=σAT4         QA=σ.π(2×10-2)2×C481×10-28

QB=σ.π(4×10-2)2×C4256×10-28

and  QC=σ.π(6×10-2)2×C4625×10-28

Hence QB>QC>QA



Q 13 :

2 kg of ice at -20°C is mixed with 5 kg of water at 20°C in an insulating vessel having a negligible heat capacity. Calculate the final mass of water remaining in the container. It is given that the specific heats of water & ice are 1 kcal/kg/C and 0.5 kcal/kg/C while the latent heat of fusion of ice is 80 kcal/kg                   [2003]

  • 7 kg

     

  • 6 kg

     

  • 4 kg

     

  • 2 kg

     

(2)

Heat released when 5 kg of water at 20°C falls to 0°C

temperature=mCwΔT=5×1×20=100 kcal

Heat required by 2kg ice at -20°C to convert into 2kg of ice at 0°C=mCiceΔT=2×0.5×20=20 kcal

Hence remaining 100-20=80 kcal of heat will melt m kg ice at 0°C to water at 0°C

Q=mL80=m×80m=1 kg

Therefore the amount of water at 0°C=5 kg+1 kg=6 kg

And amount of ice at 0°C=2-1=1 kg



Q 14 :

The graph, shown in the adjacent diagram, represents the variation of temperature (T) of two bodies, x and y having same surface area, with time (t) due to the emission of radiation. Find the correct relation between the emissivity and absorptivity power of the two bodies                     [2003]

[IMAGE 422]

  • Ex>Ey & ax<ay

     

  • Ex<Ey & ax>ay

     

  • Ex>Ey & ax>ay

     

  • Ex<Ey & ax<ay

     

(3)

[IMAGE 423]

From the graph,

(-dTdt)x>(-dTdt)y

Rate of cooling,

(-dTdt)emissivity (e)

  Ex>Ey

Also ax>ay as good absorbers are good emitters.



Q 15 :

An ideal Black-body at room temperature is thrown into a furnace. It is observed that          [2002]

  • initially it is the darkest body and at later times the brightest

     

  • it is the darkest body at all times

     

  • it cannot be distinguished at all times

     

  • initially it is the darkest body and at later times it cannot be distinguished

     

(1)

According to Kirchhoff's law, good absorbers are good emitters and bad reflectors.

At high temperature (in the furnace), since it absorbs more energy, it emits more radiations as well and hence is the brightest and it is the darkest body initially.



Q 16 :

Three rods made of same material and having the same cross-section have been joined as shown in the figure. Each rod is of the same length. The left and right ends are kept at 0°C and 90°C respectively. The temperature of the junction of the three rods will be                     [2001]

[IMAGE 424]

  • 45°C

     

  • 60°C

     

  • 30°C

     

  • 20°C

     

(2)

[IMAGE 425]

Let θ°C be the temperature of junction at B. Let Q is the heat flowing per second from A at 90°C to B at θ°C on account of temperature difference.

  Q=KA(90-θ)l                           ...(i)

And same for C to B.

Q=KA(90-θ)l

    The heat flowing per second from B to D

  2Q=KA(θ-0)l                              ...(ii)

Dividing eq. (ii) by (i)

2=θ90-θθ=60°

Hence temperature of the junction θ=60°C



Q 17 :

The plots of intensity versus wavelength for three black bodies at temperature T1, T2 and T3 respectively are as shown. Their temperatures are such that             [2000]

[IMAGE 426]

  • T1>T2>T3

     

  • T1>T3>T2

     

  • T2>T3>T1

     

  • T3>T2>T1

     

(2)

According to Wien's displacement law, λT=constant

From graph λ1<λ3<λ2             T1>T3>T2



Q 18 :

A block of ice at -10°C is slowly heated and converted to steam at 100°C. Which of the following curves represents the phenomenon qualitatively?           [2000]

  • [IMAGE 427]

     

  • [IMAGE 428]

     

  • [IMAGE 429]

     

  • [IMAGE 430]

     

(1)

O → A, the temperature of ice changes from −10°C to 0°C.
A → B, ice at 0°C melts into water at 0°C.
B → C, water at 0°C changes into water at 100°C.
C → D, water at 100°C changes into steam at 100°C.

[IMAGE 431]



Q 19 :

A small object is placed at the center of a large evacuated hollow spherical container. Assume that the container is maintained at 0 K. At time t=0, the temperature of the object is 200 K. The temperature of the object becomes 100 K at t=t1 and 50 K at t=t2. Assume the object and the container to be ideal black bodies. The heat capacity of the object does not depend on temperature. The ratio (t2t1) is _____________.                             [2021]



(9)

 Heat radiated =σAT4=-msdTdt

So,  200100dTT4=0t1kdt13T3|200100=kt1

or,     13(11003-12003)=kt1                        ...(i)

Similarly,    13T3|20050=kt2

13(1503-12003)=kt2                 ...(ii)

Dividing eq. (ii) by (i),

t2t1=(2003-5032003-1003)1003503=9



Q 20 :

A metal is heated in a furnace where a sensor is kept above the metal surface to read the power radiated (P) by the metal. The sensor has a scale that displays log2(PP0), where P0 is a constant. When the metal surface is at a temperature of 487°C, the sensor shows a value 1. Assume that the emissivity of the metallic surface remains constant. What is the value displayed by the sensor when the temperature of the metal surface is raised to 2767°C?                                 [2016]



(9)

According to Stefan's law,   PT4      or       P=P0T4

  log2P=log2P0+log2T4             log2PP0=4log2T

At   T=487°C=760K,   log2PP0=4log2760=1          ...(i)

At  T=2767°C=3040K,

log2PP0=4log23040=4log2(760×4)

               =4[log2760+log222]=4log2760+8=1+8=9



Q 21 :

Two spherical stars A and B emit blackbody radiation. The radius of A is 400 times that of B and A emits 104 times the power emitted from B. The ratio (λAλB) of their wavelengths λA and λB at which the peaks occur in their respective radiation curves is                       [2015]



(2)

 From (i) Stefan-Boltzmann law,  P=σAT4  and (ii) Wien's displacement law, λm×T=constant

PAPB=AAABTA4TB4=AAAB×λB4λA4

  λAλB=[AAAB×PBPA]14=[RA2RB2×PBPA]14=[400×400104]14

  λAλB=2



Q 22 :

Two spherical bodies A (radius 6 cm) and B (radius 18 cm) are at temperature T1 and T2, respectively. The maximum intensity in the emission spectrum of A is at 500 nm and in that of B is at 1500 nm. Considering them to be black bodies, what will be the ratio of the rate of total energy radiated by A to that of B?                 [2010]



(9)

According to Stefan-Boltzmann law,

Rate of total energy radiated by ARate of total energy radiated by B

=σT14(4πr12)σT24(4πr22)=(T1T2)4×(r1r2)2

=(λm2λm1)4(r1r2)2    [T1T2=λm2λm1 by Wien's law]

=(1500500)4(618)2=9



Q 23 :

A piece of ice (heat capacity =2100 Jkg-1°C-1 and latent heat =3.36×105 Jkg-1) of mass m grams is at -5°C at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally when the ice-water mixture is in equilibrium, it is found that 1 gm of ice has melted. Assuming there is no other heat exchange in the process, the value of m is                    [2010]



(8)

As there is no other heat exchange in this process, So, Heat supplied = Heat used in converting m grams of ice from -5°C to 0°C + Heat used in converting 1 gram of ice at 0°C to water at 0°C

or, 420=mcΔθ+ml

420=m×21001000×5+1×3.36×1051000

420=m×10.5+336

  m=8410.5=8 g



Q 24 :

Two identical plates P and Q, radiating as perfect black bodies, are kept in vacuum at constant absolute temperatures TP and TQ, respectively, with TQ<TP, as shown in Fig. 1. The radiated power transferred per unit area from P to Q is W0. Subsequently, two more plates, identical to P and Q, are introduced between P and Q, as shown in Fig. 2. Assume that heat transfer takes place only between adjacent plates. If the power transferred per unit area in the direction from P to Q (Fig. 2) in the steady state is WS, then the ratio W0WS is ______.                       [2025]

[IMAGE 432]



(3)

Initially the radiated power transferred per unit area from P to Q

[IMAGE 433]

Finally in steady state

σ(TP4-T14)=σ(T14-T24)

σ(T14-T24)=σ(T24-TQ4)

[IMAGE 434]

Adding we get

TP4-T14=T24-TQ4T14+T24=TP4+TQ4

or, T14-T24=TP4-T14

Adding: T14=2TP4+TQ43

So, WS=σ(TP4-T14)

=σ[TP4-(2TP4+TQ43)]=σ(TP4-TQ43)

  W0WS=σ(TP4-TQ4)σ(TP4-TQ43)=3



Q 25 :

The specific heat capacity of a substance is temperature dependent and is given by the formula C=kT, where k is a constant of suitable dimensions in SI units, and T is the absolute temperature. If the heat required to raise the temperature of 1 kg of the substance from -73°C to 27°C is nk, the value of n is _____.   [Given: 0 K=-273°C.]                            [2024]



(25000)

Given, mass m=1 kg,

Ti=-73°C=200 K

Tf=27°C=300 K

Q=mcdT=1·kTdT

Q=2003001·kTdT=k200300TdT

=k2[T2]200300=k2[3002-2002]=25000K

Hence n=25000



Q 26 :

A liquid at 30°C is poured very slowly into a calorimeter that is at temperature of 110°C. The boiling temperature of the liquid is 80°C. It is found that the first 5 gm of the liquid completely evaporates. After pouring another 80 gm of the liquid, its specific heat will be ____ °C.

[Neglect the heat exchange with surrounding]                                    [2019]



(270)

Let C be the specific heat capacity of liquid and L be the latent heat of vapourisation.

From principle of calorimetry,

Heat lost=heat gain

mcScΔT=mCΔT+mL

or  mcSc(110-80)=5C(80-30)+5L                  ...(i)

Where, mc=mass of calorimeter

Sc=sp. heat of calorimeter

Again, when 80g liquid is poured and equilibrium temperature is 50°C

mcSc(80-50)=80C(50-30)                   ...(ii)

From eq. (i) & (ii)

1600C=250C+5L

  LC=13505=270°C



Q 27 :

Two conducting cylinders of equal length but different radii are connected in series between two heat baths kept at temperatures T1=300 K and T2=100 K, as shown in the figure. The radius of the bigger cylinder is twice that of the smaller one and the thermal conductivities of the materials of the smaller and the larger cylinders are K1 and K2, respectively. If the temperature at the junction of the two cylinders in the steady state is 200 K, then K1K2 = ________ .               [2018]

[IMAGE 435]



(4)

Rate of heat flow will be same,

Rate of heat flow dQdt=temp. differencethermal resistance=1R(T2-T1)

where R=LKA

300-200R1=200-100R2  or  R1=R2

L1K1A1=L2K2A2       K1K2=A2A1=π(2r)2πr2=4    [L1=L2=L]



Q 28 :

A composite block is made of slabs A, B, C, D and E of different thermal conductivities (given in terms of a constant K and sizes (given in terms of length, L)) as shown in the figure. All slabs are of same width. Heat ‘Q’ flows only from left to right through the blocks. Then in steady state               [2011]

[IMAGE 436]

  • heat flow through A and E slabs are same.

     

  • heat flow through slab E is maximum.

     

  • temperature difference across slab E is smallest.

     

  • heat flow through C = heat flow through B + heat flow through D.

     

Select one or more options

(1, 3, 4)

According to question, heat Q flows only from left to right through the blocks. Hence heat flow through slab A and E are the same.

[IMAGE 437]

We know that thermal resistance R=KA

Let the width of slabs be W. Then

RA=L2K(4L)W=18KW,    RB=4L3K(LW)=43KW

RC=4L4K(2LW)=12KW,    RD=4L5K(LW)=45KW

RE=L6K(4LW)=124KW

Now, ΔT=QR

Since the resistance to heat flow is least for slab E, the temperature difference across E is smallest.

Also

QC=ΔTCRC=ΔTC1/2KW=2KW(ΔTC)

QB=ΔTBRB=ΔTC4/3KW=3KW(ΔTC)4        [ΔTB=ΔTC]

QD=ΔTDRD=ΔTC4/5KW=5KW(ΔTC)4    [ΔTD=ΔTC]

QB+QD=3KW(ΔTC)4+5KW(ΔTC)4

=8KW(ΔTC)4=2KW(ΔTC)=QC

i.e., QC=QB+QD



Q 29 :

A black body of temperature T is inside chamber of T0 temperature initially. Sun rays are allowed to fall from a hole in the top of chamber. If the temperature of black body (T) and chamber (T0) remains constant, then                             [2006]

[IMAGE 438]

  • Black body will absorb more radiation

     

  • Black body will absorb less radiation

     

  • Black body emit more energy

     

  • Black body emit energy equal to energy absorbed by it

     

(4)

Since the temperature of black body and chamber remains constant so energy emitted = energy absorbed by black body.