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Question #77078

Difficulty: 8/10 | 🟢 Free

The Question

A solid sample of mass mm and unknown specific heat cc is heated to an initial measured temperature T1′T_1'. It is then dropped into a calorimeter containing a liquid, with the system having a combined heat capacity SS. The initial measured temperature of the calorimeter and the surrounding environment is T2′T_2'. The mixture reaches a maximum measured temperature T3′T_3' after a time tt. During this mixing phase, heat is lost to the surroundings according to Newton's Law of Cooling, with a cooling constant kk (in J⋅K−1⋅s−1\text{J} \cdot \text{K}^{-1} \cdot \text{s}^{-1}). All temperatures are measured using a thermometer whose scale is printed on a metal strip with a coefficient of linear expansion α\alpha. The thermometer is perfectly calibrated at 0∘C0^\circ\text{C}, and the true temperature TT is strictly proportional to the absolute length of the thermometric liquid column. Assuming α\alpha is very small such that terms of order α2\alpha^2 and higher can be neglected, determine the specific heat cc of the sample.

Options

  • A.

    2S+kt2m(T3′−T2′T1′−T3′)[1−α(T1′−T2′)]\frac{2S + k t}{2m} \left( \frac{T_3' - T_2'}{T_1' - T_3'} \right) [1 - \alpha(T_1' - T_2')]

  • B.

    2S+kt2m(T3′−T2′T1′−T3′)[1−α(T1′+T2′−2T3′)]\frac{2S + k t}{2m} \left( \frac{T_3' - T_2'}{T_1' - T_3'} \right) [1 - \alpha(T_1' + T_2' - 2T_3')]

  • C.

    S+ktm(T3′−T2′T1′−T3′)[1−α(T1′−T2′)]\frac{S + k t}{m} \left( \frac{T_3' - T_2'}{T_1' - T_3'} \right) [1 - \alpha(T_1' - T_2')]

  • D.

    2S+kt2m(T3′−T2′T1′−T3′)[1+α(T1′−T2′)]\frac{2S + k t}{2m} \left( \frac{T_3' - T_2'}{T_1' - T_3'} \right) [1 + \alpha(T_1' - T_2')]

Check Your Answer

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

Examiner Levers

Visualization Stress

Requires mental rotation or 3D mapping (Optics/Geometry/Vectors).

Difficulty Boost : +4

Variable Manipulation

Algebraic solutions only (No numbers); tests pure symbolic logic.

Difficulty Boost : +4

Common Mistakes

This question tests two distinct, nested physical traps.

First, students often misapply the thermal expansion correction by assuming the error scales with the temperature difference directly (i.e., ΔT=ΔT′(1+αΔT′)\Delta T = \Delta T'(1 + \alpha \Delta T')), which leads to the incorrect α(T1′+T2′−2T3′)\alpha(T_1' + T_2' - 2T_3') term in Option B.

The correct approach requires expanding the squares of individual temperatures, revealing that the final mixture temperature T3′T_3' elegantly cancels out of the correction factor.

Second, it tests the application of Newton's Law of Cooling during a dynamic mixing phase; students who fail to use the average temperature difference (T3−T2)/2(T_3 - T_2)/2 will incorrectly use the maximum difference, leading to the unhalved ktkt term in Option C.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterUnits and Measurements - Class 11 Part 1
ConceptZero Error Correction in Instruments
Difficulty8
Cognitive Load8
Target Solve Time180 sec
AccessFree