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Question #42015

Difficulty: 8/10 | 🟢 Free

The Question

A solid of mass MM is heated to a temperature measured as θH\theta_H by a thermometer and dropped into a calorimeter of combined heat capacity CC (including water), initially at a measured temperature of 0∘C0^\circ\text{C}. The final steady temperature of the mixture, measured by the same thermometer, is θ\theta. The thermometer's scale is calibrated correctly at 0∘C0^\circ\text{C} but has a coefficient of linear expansion α\alpha. During the mixing, the system loses heat to the surroundings (at 0∘C0^\circ\text{C}) given by Qloss=λTfQ_{\text{loss}} = \lambda T_f, where TfT_f is the true final temperature and λ\lambda is a constant. Assuming α\alpha is very small, which expression gives the specific heat capacity SS of the solid, correct to first order in α\alpha?

Options

  • A.

    (C+λ)θM(θH−θ)(1−αθH)\frac{(C + \lambda) \theta}{M (\theta_H - \theta)} (1 - \alpha \theta_H)

  • B.

    (C+λ)θM(θH−θ)(1+αθH)\frac{(C + \lambda) \theta}{M (\theta_H - \theta)} (1 + \alpha \theta_H)

  • C.

    (C+λ)θM(θH−θ)(1−α(θH−θ))\frac{(C + \lambda) \theta}{M (\theta_H - \theta)} (1 - \alpha (\theta_H - \theta))

  • D.

    (C−λ)θM(θH−θ)(1−αθH)\frac{(C - \lambda) \theta}{M (\theta_H - \theta)} (1 - \alpha \theta_H)

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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

Zero Error Correction|Calorimetry with Heat Loss|Thermal Expansion of Area and Volume|Binomial Approximation

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to synthesize instrumental error with thermodynamic energy conservation.

The trap lies in the 'double correction': students must first recognize that thermal expansion of the thermometer scale causes a temperature-dependent zero error, requiring the transformation of measured temperatures to true temperatures (T≈θ(1+αθ)T \approx \theta(1+\alpha\theta)).

Secondly, they must correctly incorporate the non-conservative heat loss term into the calorimetric balance.

A common mistake is applying the binomial approximation incorrectly or failing to realize that the scale's expansion leads to a lower reading, causing sign errors in the correction factor.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterUnits and Measurements - Class 11 Part 1
ConceptZero Error Correction in Instruments
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree