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Question #42142

Difficulty: 8/10 | 🟢 Free

The Question

A cylindrical heating wire of initial resistance R0=1 ΩR_0 = 1 \text{ }\Omega at T0=300 KT_0 = 300 \text{ K} is submerged in an insulating vessel containing m=1 kgm = 1 \text{ kg} of a liquid with specific heat capacity s=4000 J kg−1K−1s = 4000 \text{ J kg}^{-1} \text{K}^{-1}. A constant voltage V=20 VV = 20 \text{ V} is applied across the wire. The temperature coefficient of resistivity of the wire is αρ=5×10−3 K−1\alpha_\rho = 5 \times 10^{-3} \text{ K}^{-1} and its coefficient of linear thermal expansion is αL=1×10−3 K−1\alpha_L = 1 \times 10^{-3} \text{ K}^{-1}. Assuming the heat capacity of the wire and the vessel are negligible, and no heat is lost to the surroundings, find the time tt required to heat the liquid to T=350 KT = 350 \text{ K}. (Use the approximation (1+x)n≈1+nx(1+x)^n \approx 1+nx for small xx).

Options

  • A.

    550 s550 \text{ s}

  • B.

    562.5 s562.5 \text{ s}

  • C.

    500 s500 \text{ s}

  • D.

    575 s575 \text{ s}

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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

Specific Heat Capacity of Phase Mixtures|Temperature Coefficient (Alpha)|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to couple Joule heating with thermodynamics while accounting for non-linear feedback and geometric thermal expansion.

A common trap is assuming constant power P=V2/R0P = V^2/R_0, which ignores the temperature dependence of resistance and leads to a simple linear time relation (yielding 500 s).

The secondary, more advanced trap is forgetting that resistance R=ρL/AR = \rho L / A depends on the physical dimensions of the wire, which also expand.

Students often use only the resistivity coefficient αρ\alpha_\rho (yielding 562.5 s) or incorrectly add the linear expansion coefficient instead of deriving the effective coefficient αeff=αρ−αL\alpha_{eff} = \alpha_\rho - \alpha_L.

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Question Information

SubjectPhysics
ChapterThermal Properties of Matter - Class 11 Part 2
ConceptSpecific Heat Capacity of Phase Mixtures
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree