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Thermal Properties of Matter

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Question
CBSEENPH11020788
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A black body at 1227°C emits radiations with maximum intensity at a wavelength of 5000 Å. If the temperature of the body is increased by 1000°C, the maximum intensity will be observed at

  • 4000 Å

  • 5000 Å

  • 6000 Å

  • 3000 Å

Solution

D.

3000 Å

The product of wavelength corresponding to maximum intensity of radiation and temperature of the body in Kelvin is constant.
According to Wien's law
   straight lambda subscript straight m straight T space equals space constant space left parenthesis say space straight b right parenthesis
where straight lambda subscript straight m is wavelength corresponding to maximum intensity of radiation and T is temperature of the body in Kelvin.
therefore space space space space straight lambda subscript straight m apostrophe end subscript over straight lambda subscript straight m space equals space fraction numerator straight T over denominator straight T apostrophe end fraction
Given,   T = 1227 + 273 = 1500 K,
             T' = 1227 + 1000 + 273 = 2500 K
              straight lambda subscript straight m space equals space 5000 space straight Å Hence comma space space space straight lambda subscript straight m apostrophe end subscript space equals space 1500 over 2500 cross times 5000 space equals space 3000 space straight Å

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Question
CBSEENPH11020777
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A black body is at 727o C. It emits energy at a rate which is proportional to:

  • (727)2

  • (1000)4

  • (1000)2

  • (727)2

Solution

B.

(1000)4

Amount of heat energy radiated per secnd by unit area of black body is directly proportional to fourth power of absoulte temperature.
Accordign to steafan's law,
straight E space proportional to straight T to the power of 4 space straight E thin space equals σT to the power of 4
Where σ is constatn of proportionality and called Steafan's constant. Its value is
5.67 x 10-8 Wm-2 K-4.
hence, E = (727+273)4
E = (1000)4

Question
CBSEENPH11020725
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A block body t 227o C radiates heat at the rate of 7 cal cm-2 s-1. At  a temperature of 727o C, the rate of heat radiated in the same units will be 

  • 60

  • 50

  • 112

  • 80

Solution

C.

112

Apply Stefan's law of radiation.
straight E space equals σT to the power of 4 straight E subscript 1 over straight E subscript 2 space equals space open square brackets straight T subscript 1 over straight T subscript 2 close square brackets to the power of 4 straight E subscript 2 space equals space 7 space open square brackets fraction numerator 273 plus 727 over denominator 273 plus 227 end fraction close square brackets to the power of 4 space equals open parentheses 1000 over 500 close parentheses to the power of 4 space straight x space 7 equals space 112 space cal space cm to the power of negative 2 end exponent space straight s to the power of negative 1 end exponent

Question
CBSEENPH11020685
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A cylindrical metallic rod in thermal contact with two reservoirs of heat at its two ends conducts an amount of heat Q in time t. The metallic rod is melted and the material is formed into a rod of half the radius of the original rod. What is the amount of heat conducted by the new rod when placed in thermal contact with the two reservoirs in time t?

  • Q/4

  • Q/16

  • 2Q

  • Q/2

Solution

B.

Q/16

In steady state, the amount of heat flowing from one face to the other face in time t is given by 

straight Q space equals space fraction numerator KA space left parenthesis straight theta subscript 1 minus straight theta subscript 2 right parenthesis straight t over denominator straight l end fraction space where space straight K space is space coefficient space of space thermal space conductivity space of space material space of space rod rightwards double arrow space straight Q over straight t space proportional to space straight A over straight l space proportional to space straight r squared over straight I space space.. space left parenthesis straight i right parenthesis As space the space metallic space rod space is space melted space and space the space material space is space formed space into space straight a space rod space of space half space the space radius space straight V subscript 1 space equals space straight V subscript 2 πr subscript 1 superscript 2 straight l subscript 1 space equals space πr subscript 2 superscript 2 straight l subscript 2 straight l subscript 1 space equals space straight l subscript 2 over 4 space space space... space left parenthesis iI right parenthesis Now comma space from space eqs. space left parenthesis straight i right parenthesis thin space and space left parenthesis ii right parenthesis space fraction numerator straight Q subscript 1 over denominator straight Q subscript 2 space end fraction space equals space fraction numerator straight r subscript 1 superscript 2 over denominator straight l subscript 1 end fraction space straight x space fraction numerator straight l subscript 2 over denominator straight r subscript 2 superscript 2 end fraction space equals space fraction numerator straight r subscript 1 superscript 2 over denominator straight l subscript 1 end fraction space straight x space fraction numerator 4 straight l subscript 1 over denominator left parenthesis straight r subscript 1 divided by 2 right parenthesis squared end fraction space rightwards double arrow space straight Q subscript 1 space equals space 16 space straight Q subscript 2

Question
CBSEENPH11020809
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A gas mixture consists of 2 moles of O2 and 4 moles of Ar at temperature T. Neglecting all vibrational modes, the total internal energy of the system is

  • 4 RT

  • 15 RT

  • 9 RT

  • 11 RT

Solution

D.

11 RT

straight U space equals space straight n subscript 1 straight f subscript 1 over 2 RT space plus space straight n subscript 2 straight f subscript 2 over 2 RT equals space 2 space straight x space 5 over 2 space RT space plus space 4 space straight x space 3 over 2 space RT space equals space 5 RT space plus space 6 RT straight U space equals space 11 space RT

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