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Heat and Thermodynamics question

2023 · 30 Jan · Shift 2 · Q63
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  5. /2023 · 30 Jan · Shift 2 · Q63

Heat and Thermodynamics question

2023 · 30 Jan · Shift 2 · Q63

JEE MainPhysicsHeat and ThermodynamicsNumerical+4 / −1
A faulty thermometer reads 5∘C5^{\circ} \mathrm{C}5∘C in melting ice and 95∘C95^{\circ} \mathrm{C}95∘C in stream. The correct temperature on absolute scale will be ‾\underline{\hspace{2cm}}​K\mathrm{K}K when the faulty thermometer reads 41∘C41^{\circ} \mathrm{C}41∘C.
Numerical answer
View written solutionFree

Correct answer: 313

  1. Let the faulty thermometer reading be FFF and the true temperature in degree Celsius be ttt.

  2. Since the thermometer is linear:

    • At melting ice: true temperature =0∘C=0^\circ C=0∘C, faulty reading =5∘C=5^\circ C=5∘C
    • At steam point: true temperature =100∘C=100^\circ C=100∘C, faulty reading =95∘C=95^\circ C=95∘C

    So the linear relation between faulty reading and true temperature is F−595−5=t−0100−0\frac{F-5}{95-5}=\frac{t-0}{100-0}95−5F−5​=100−0t−0​

  3. Substitute F=41∘CF=41^\circ CF=41∘C: 41−590=t100\frac{41-5}{90}=\frac{t}{100}9041−5​=100t​ 3690=t100\frac{36}{90}=\frac{t}{100}9036​=100t​ t=100×3690=40∘Ct=100\times \frac{36}{90}=40^\circ Ct=100×9036​=40∘C

  4. Convert to absolute scale: T=273+40=313 KT = 273 + 40 = 313\,\text{K}T=273+40=313K

  5. Therefore, the correct temperature is 313 K\boxed{313\,\text{K}}313K​

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