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Question #43154

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The Question

A solid body of heat capacity $C$ (in $\text{J/K}$ ) is placed in a room maintained at a constant temperature $T_0$ . The body is heated by an internal electrical heater that supplies power according to the relation $P(t) = P_0(1 + \sin(\omega t))$ , where $P_0$ and $\omega$ are positive constants. The body loses heat to the surroundings strictly according to Newton's law of cooling, with a cooling constant $K$ (in $\text{W/K}$ ). After a sufficiently long time, the temperature of the body reaches a steady oscillatory state. Which of the following expressions represents the maximum temperature reached by the body in this steady state?

Options

A.
$T_0 + \frac{2P_0}{K}$
B.
$T_0 + \frac{P_0}{K} + \frac{P_0}{\omega C}$
C.
$T_0 + \frac{P_0}{K} + \frac{P_0}{\sqrt{K^2 + \omega^2 C^2}}$
D.
$T_0 + \frac{P_0}{\sqrt{K^2 + \omega^2 C^2}}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Average Power over Cycle|Calorimetry with Heat Loss|Differential Equations

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to analyze transient and steady-state thermodynamics using an electrical-thermal analogy.

The primary trap lies in misapplying steady-state calorimetry.

Many students will incorrectly assume that the maximum temperature corresponds directly to the maximum instantaneous power input ( $2P_0$ ), leading to $T_{max} = T_0 + 2P_0/K$ .

This ignores the thermal inertia (heat capacity $C$ ) of the body, which causes a phase lag and damps the amplitude of the temperature oscillations.

Another common mistake is treating the oscillatory heating as if there is no simultaneous cooling, leading to an incorrect amplitude of $P_0/(\omega C)$ .

The correct approach requires setting up the differential equation for Newton's Law of Cooling with a time-varying source and recognizing that the steady-state solution is a superposition of a shifted mean temperature and a damped oscillatory component.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Alternating Current - Class 12
Concept	Average Power over Cycle
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free