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Question #42147

Difficulty: 8/10 | 🟢 Free

The Question

A heating coil is submerged in a calorimeter containing a liquid. The coil is connected to a constant voltage source of V=20 VV = 20\ \text{V}. The resistance of the coil at the surrounding temperature Ts=25∘CT_s = 25^\circ\text{C} is R0=10 ΩR_0 = 10\ \Omega, and its temperature coefficient of resistance is α=0.05 ∘C−1\alpha = 0.05\ ^\circ\text{C}^{-1}. The calorimeter loses heat to the surroundings according to Newton's law of cooling, with a heat transfer coefficient k=1 W/∘Ck = 1\ \text{W/}^\circ\text{C}. Assuming the liquid is continuously stirred and maintains a uniform temperature, what is the steady-state temperature of the liquid?

Options

  • A.

    45∘C45^\circ\text{C}

  • B.

    65∘C65^\circ\text{C}

  • C.

    −15∘C-15^\circ\text{C}

  • D.

    55∘C55^\circ\text{C}

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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|Internal Resistance Variation|Current Electricity

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to model a non-linear thermal-electric feedback loop.

A common trap is assuming the power dissipated by the coil remains constant (P=V2/R0P = V^2/R_0), which ignores the temperature dependence of the internal resistance.

By doing so, students incorrectly calculate a linear steady-state condition (yielding 65∘C65^\circ\text{C}), failing to recognize that as the temperature rises, the resistance increases, thereby reducing the Joule heating rate and leading to a lower actual steady-state temperature.

Another trap is incorrectly using P=I2RP = I^2R while assuming constant current, which yields a physically absurd negative temperature change.

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