Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Heat and Thermodynamics question

2010 · Shift 1 · Q63
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Advanced
  3. /Physics
  4. /Heat and Thermodynamics
  5. /2010 · Shift 1 · Q63

Heat and Thermodynamics question

2010 · Shift 1 · Q63

JEE AdvancedPhysicsHeat and ThermodynamicsNumerical+3 / −1
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?
Numerical answer
View written solutionFree

Correct answer: 9

The user wants to find the ratio of the rate of total energy radiated by two spherical black bodies, A and B. This requires using both Wien's Displacement Law and the Stefan-Boltzmann Law.

Step 1: Find the ratio of the temperatures of the two bodies.

Wien's Displacement Law states that the wavelength of maximum emission, λmax\lambda_{max}λmax​, is inversely proportional to the absolute temperature, T, of a black body: λmaxT=b\lambda_{max} T = bλmax​T=b where bbb is Wien's constant.

For body A: λmax,AT1=b\lambda_{max, A} T_1 = bλmax,A​T1​=b For body B: λmax,BT2=b\lambda_{max, B} T_2 = bλmax,B​T2​=b

From these two equations, we can find the ratio of the temperatures: λmax,AT1=λmax,BT2\lambda_{max, A} T_1 = \lambda_{max, B} T_2λmax,A​T1​=λmax,B​T2​ T1T2=λmax,Bλmax,A\frac{T_1}{T_2} = \frac{\lambda_{max, B}}{\lambda_{max, A}}T2​T1​​=λmax,A​λmax,B​​

Given λmax,A=500 nm\lambda_{max, A} = 500 \text{ nm}λmax,A​=500 nm and λmax,B=1500 nm\lambda_{max, B} = 1500 \text{ nm}λmax,B​=1500 nm: T1T2=1500 nm500 nm=3\frac{T_1}{T_2} = \frac{1500 \text{ nm}}{500 \text{ nm}} = 3T2​T1​​=500 nm1500 nm​=3

Step 2: Find the ratio of the surface areas of the two bodies.

The bodies are spherical, so the surface area is given by the formula A=4πr2A = 4 \pi r^2A=4πr2. The ratio of the surface areas is: AAAB=4πrA24πrB2=(rArB)2\frac{A_A}{A_B} = \frac{4 \pi r_A^2}{4 \pi r_B^2} = \left(\frac{r_A}{r_B}\right)^2AB​AA​​=4πrB2​4πrA2​​=(rB​rA​​)2

Given the radii rA=6 cmr_A = 6 \text{ cm}rA​=6 cm and rB=18 cmr_B = 18 \text{ cm}rB​=18 cm: AAAB=(6 cm18 cm)2=(13)2=19\frac{A_A}{A_B} = \left(\frac{6 \text{ cm}}{18 \text{ cm}}\right)^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}AB​AA​​=(18 cm6 cm​)2=(31​)2=91​

Step 3: Find the ratio of the rate of total energy radiated.

The Stefan-Boltzmann Law states that the total energy radiated per unit time (power, P) by a black body is given by: P=σAT4P = \sigma A T^4P=σAT4 where σ\sigmaσ is the Stefan-Boltzmann constant, A is the surface area, and T is the absolute temperature.

The ratio of the power radiated by A to that of B is: PAPB=σAAT14σABT24=(AAAB)(T1T2)4\frac{P_A}{P_B} = \frac{\sigma A_A T_1^4}{\sigma A_B T_2^4} = \left(\frac{A_A}{A_B}\right) \left(\frac{T_1}{T_2}\right)^4PB​PA​​=σAB​T24​σAA​T14​​=(AB​AA​​)(T2​T1​​)4

Step 4: Calculate the final ratio.

Substitute the ratios for area and temperature calculated in the previous steps: PAPB=(19)(3)4\frac{P_A}{P_B} = \left(\frac{1}{9}\right) (3)^4PB​PA​​=(91​)(3)4 PAPB=19×81\frac{P_A}{P_B} = \frac{1}{9} \times 81PB​PA​​=91​×81 PAPB=9\frac{P_A}{P_B} = 9PB​PA​​=9

The ratio of the rate of total energy radiated by A to that of B is 9.

PreviousNext

More from Heat and Thermodynamics

  • A piece of ice (heat capacity = 2100 J kg-1 oC-1 and latent heat = 3.36 × 105 J kg-1 ) of mass m grams is at - 5 oC at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally when the ice-water…2010 · Numerical
  • One mole of an ideal gas in initial state A undergoes a cyclic process ABCA, as shown in the figure. Its pressure at A is P0. Choose the correct option(s) from the following: Includes diagram2010 · Multiple correct
  • A diatomic ideal gas is compressed adiabatically 321​ of its initial volume. If the initial temperature of the gas is Ti (in Kelvin) and the final temperature is a Ti, the value of a is2010 · Numerical
  • CV​ and CP​ denote the molar specific heat capacities of a gas at constant volume and constant pressure, respectively. Then2009 · Multiple correct
  • The figure shows the PV plot of an ideal gas taken through a cycle ABCDA. The part ABC is a semicircle and CDA is half of an ellipse. Then, Includes diagram2009 · Multiple correct
  • Column II gives certain systems undergoing a process. Column I suggests changes in some of the parameters related to the system. Match the statements in Column I to the appropriate process(es) from Column II: Includes table2009 · MCQ
  • A metal rod AB of length 10x has its one end A in ice at 0 ∘ C and the other end B in water at 100 ∘ C. If a point P on the rod is maintained at 400 ∘ C, then it is found that equal amounts of water and ice evaporate…2009 · Numerical
  • An ideal gas is expanding such that PT 2 = constant. The coefficient of volume expansion of the gas is2008 · MCQ