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A planet of radius R is made of uniform material that generates power P uniformly via radioactive decay. Heat flows radially outward by thermal conduction (thermal conductivity k). The temperature difference between the planet's centre and its surface equals P / (alpha * pi * k * R). Find the value of alpha.
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Correct answer: 4
Solution
Power generated per unit volume: q = P / (4*pi*R³/3) = 3P/(4*pi*R³). At radius r, total power from inner sphere = q*(4*pi*r³/3) = P*r³/R³. By Fourier's law: this power flows outward, so -k*(4*pi*r²)*(dT/dr) = P*r³/R³. Thus dT/dr = -P*r / (4*pi*k*R³). Integrating from centre (T_c) to surface (Tₛ): T_c - Tₛ = integral₀^R [P*r/(4*pi*k*R³)] dr = P/(4*pi*k*R³) * [r²/2]₀^R = P/(4*pi*k*R³) * R²/2 = P/(8*pi*k*R). So delta_T = P/(8*pi*k*R). Comparing with P/(alpha*pi*k*R): alpha = 8.
Related JEE Advanced Physics questions
- Match the temperature of a blackbody listed in Group-I to the corresponding statement in Group-II, and select the correct answer.
[Given: Wien’s constant = 2.9 × 10⁻³ m-K and hc/e = 1.24 × 10⁻⁶ V-m]
Group-I:
(P) 2000 K
(Q) 3000 K
(R) 5000 K
(S) 10000 K
Group-II:
(1) The peak wavelength of emitted radiation can cause photoelectron ejection from a metal with a work function of 4 eV.
(2) The peak wavelength of emitted radiation falls within the visible spectrum.
(3) The peak wavelength of emitted radiation produces the broadest central diffraction maximum in a single-slit setup.
(4) The energy radiated per unit area is one-sixteenth of that emitted by a blackbody at 6000 K.
(5) The peak wavelength of emitted radiation is suitable for imaging human bones.
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- A continuous-flow calorimeter is used in two separate experiments to find the specific heat of a liquid. In the first trial, supplying 60 W raises the liquid temperature by 10 K. In the second trial, the power is doubled to 120 W, but the same 10 K rise is maintained by tripling the flow rate. Assuming heat loss to surroundings is the same in both trials, what is the power lost to the surroundings?
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