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Question #42149

Difficulty: 8/10 | 🟢 Free

The Question

A heating coil of resistance RR and inductance LL is immersed in a calorimeter of heat capacity CC. At t=0t=0, the coil is connected to an ideal DC voltage source V0V_0. The calorimeter loses heat to the surroundings (at constant temperature T0T_0) at a rate k(T−T0)k(T - T_0), where TT is the instantaneous temperature of the calorimeter and kk is a positive constant. Assume the electrical time constant of the coil is exactly equal to the thermal time constant of the calorimeter, i.e., LR=Ck=τ\frac{L}{R} = \frac{C}{k} = \tau. If the initial temperature of the calorimeter is T0T_0, which of the following expressions correctly gives the temperature difference ΔT(t)=T(t)−T0\Delta T(t) = T(t) - T_0 at any time t>0t > 0?

Options

  • A.

    V02kR[1−2tτe−t/τ−e−2t/τ]\frac{V_0^2}{k R} \left[ 1 - \frac{2t}{\tau} e^{-t/\tau} - e^{-2t/\tau} \right]

  • B.

    V02kR[1−e−t/τ]\frac{V_0^2}{k R} \left[ 1 - e^{-t/\tau} \right]

  • C.

    V02kR[1−(1+tτ)e−t/τ]\frac{V_0^2}{k R} \left[ 1 - \left(1 + \frac{t}{\tau}\right) e^{-t/\tau} \right]

  • D.

    V02kR[1−tτe−t/τ−e−2t/τ]\frac{V_0^2}{k R} \left[ 1 - \frac{t}{\tau} e^{-t/\tau} - e^{-2t/\tau} \right]

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Calorimetry with Heat Loss|Transient Current in RL Circuit|Work, Energy, and Power

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of transient electrical circuits and non-steady-state calorimetry.

The lethal trap lies in the non-linear coupling of the transient current to the thermal power (P∝I2P \propto I^2).

Students often incorrectly assume the power grows linearly with the current transient, leading to P∝(1−e−t/τ)P \propto (1 - e^{-t/\tau}) (Option C), or they completely ignore the transient phase and assume steady-state power V02/RV_0^2/R from t=0t=0 (Option B).

By equating the electrical and thermal time constants, the problem forces the resolution of a first-order non-homogeneous differential equation with a resonant-like forcing term, rigorously testing the ability to set up and integrate coupled physical rate equations without falling for algebraic shortcuts.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterThermal Properties of Matter - Class 11 Part 2
ConceptCalorimetry with Heat Loss
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree