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A composite slab is made of two materials with thermal conductivities K and 2K and thicknesses x and 4x respectively, placed in series. The outer surfaces are at temperatures T2 and T1 (T2 > T1). In steady state the heat-flow rate is written as f * A*(T2 - T1)*K / x. Find f.
- 1
- 1/2
- 2/3
- 1/3
Correct answer: 1/3
Solution
The series thermal resistance is x/(KA) + 2x/(KA) = 3x/(KA), so the heat rate equals (1/3)*A(T2-T1)K/x, giving f = 1/3.
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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