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Heat and Thermodynamics question

2024 · 4 Apr · Shift 1 · Q80
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Heat and Thermodynamics question

2024 · 4 Apr · Shift 1 · Q80

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
On celcius scale the temperature of body increases by 40∘C40^{\circ} \mathrm{C}40∘C. The increase in temperature on Fahrenheit scale is :
  1. A
    75∘F75^{\circ} \mathrm{F}75∘F
  2. B
    70∘F70^{\circ} \mathrm{F}70∘F
  3. C
    72∘F72^{\circ} \mathrm{F}72∘F
  4. D
    68∘F68^{\circ} \mathrm{F}68∘F
View written solutionFree

Correct answer: C

  1. Use the Celsius–Fahrenheit relation for temperature differences

    The conversion formula is F=95C+32F = \frac{9}{5}C + 32F=59​C+32

    For an increase in temperature, the constant 323232 does not affect the change. So, ΔF=95ΔC\Delta F = \frac{9}{5}\Delta CΔF=59​ΔC

  2. Substitute the given increase

    Given, ΔC=40∘C\Delta C = 40^\circ CΔC=40∘C

    Therefore, ΔF=95×40=9×8=72∘F\Delta F = \frac{9}{5} \times 40 = 9 \times 8 = 72^\circ FΔF=59​×40=9×8=72∘F

  3. Match with the options

    72∘F72^\circ F72∘F corresponds to Option C.

  4. Compare with stored correct answer

    Stored correct answer: C

    Derived answer: C

    So, they agree.

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