Start Learning Free

Before You Begin

Question #42156

Difficulty: 8/10 | 🟢 Free

The Question

A metallic sphere of mass mm is placed in a vacuum chamber maintained at absolute zero. It loses heat to the surroundings via a thermal link such that the rate of heat loss is Ploss=γTP_{loss} = \gamma T, where TT is the absolute temperature of the sphere and γ\gamma is a positive constant. Simultaneously, the sphere is subjected to a time-varying magnetic field that induces eddy currents. The power dissipated by these eddy currents is Peddy=βTP_{eddy} = \frac{\beta}{T}, where β\beta is a positive constant (this inverse dependence arises because the electrical resistivity of the metal is proportional to its absolute temperature, ρ∝T\rho \propto T). The specific heat capacity of the metal varies with temperature as s(T)=s0Ts(T) = s_0 T. If the sphere is initially at a temperature T0=2βγT_0 = 2\sqrt{\frac{\beta}{\gamma}}, what is the time tt required for it to cool down to Tf=2βγT_f = \sqrt{\frac{2\beta}{\gamma}}?

Options

  • A.

    ms0γβγ[2−2+ln⁡(2+13)]\frac{m s_0}{\gamma} \sqrt{\frac{\beta}{\gamma}} \left[ 2 - \sqrt{2} + \ln \left( \frac{\sqrt{2}+1}{\sqrt{3}} \right) \right]

  • B.

    ms0γβγln⁡3\frac{m s_0}{\gamma} \sqrt{\frac{\beta}{\gamma}} \ln 3

  • C.

    ms0γβγ[2−2−tan⁡−1(2−21+22)]\frac{m s_0}{\gamma} \sqrt{\frac{\beta}{\gamma}} \left[ 2 - \sqrt{2} - \tan^{-1} \left( \frac{2-\sqrt{2}}{1+2\sqrt{2}} \right) \right]

  • D.

    ms0γβγ[2−2−ln⁡(2+13)]\frac{m s_0}{\gamma} \sqrt{\frac{\beta}{\gamma}} \left[ 2 - \sqrt{2} - \ln \left( \frac{\sqrt{2}+1}{\sqrt{3}} \right) \right]

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Newton’s Law of Cooling Limits|Eddy Currents Dissipation|Molar Heat Capacity (Cv vs Cp)|Differential Equations

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to couple thermodynamic cooling with electrodynamic heat generation while navigating two lethal traps: a temperature-dependent internal heat source (eddy currents varying inversely with temperature due to ρ∝T\rho \propto T) and a time-varying heat capacity (s∝Ts \propto T, characteristic of the electronic specific heat in metals at low temperatures).

Students commonly make the mistake of treating resistivity or heat capacity as constants, leading to incorrect logarithmic or linear time dependencies.

Furthermore, a sign error in the energy balance equation (Pin−PoutP_{in} - P_{out}) will lead to an inverse tangent function instead of a logarithmic one, testing their fundamental understanding of the First Law of Thermodynamics in open systems.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterThermal Properties of Matter - Class 11 Part 2
ConceptNewton’s Law of Cooling Limits
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree