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Heat and Thermodynamics question

2021 · 27 Aug · Shift 2 · Q55
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Heat and Thermodynamics question

2021 · 27 Aug · Shift 2 · Q55

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
The height of victoria falls is 63 m. What is the difference in temperature of water at the top and at the bottom of fall? [Given 1 cal = 4.2 J and specific heat of water = 1 cal g −-− 1 ∘^\circ∘ 0C −-− 1]
  1. A
    0.147 ∘^\circ∘ C
  2. B
    14.76 ∘^\circ∘ C
  3. C
    1.476 ∘^\circ∘ C
  4. D
    0.014 ∘^\circ∘ C
View written solutionFree

Correct answer: A

  1. Idea: As water falls through height hhh, its gravitational potential energy converts into heat energy.

  2. Potential energy lost per gram of water

For mass m=1 g=10−3 kgm = 1\,\text{g} = 10^{-3}\,\text{kg}m=1g=10−3kg, Q=mgh=10−3×9.8×63 JQ = mgh = 10^{-3} \times 9.8 \times 63\,\text{J}Q=mgh=10−3×9.8×63J Q=0.6174 JQ = 0.6174\,\text{J}Q=0.6174J

  1. Convert this heat into calories

Given 1 cal=4.2 J1\,\text{cal} = 4.2\,\text{J}1cal=4.2J, Q=0.61744.2≈0.147 calQ = \frac{0.6174}{4.2} \approx 0.147\,\text{cal}Q=4.20.6174​≈0.147cal

  1. Use specific heat relation

Specific heat of water is c=1 cal g−1 ∘C−1c = 1\,\text{cal g}^{-1}\,^\circ\text{C}^{-1}c=1cal g−1∘C−1

For 1 g1\,\text{g}1g of water, Q=mcΔTQ = mc\Delta TQ=mcΔT 0.147=1×1×ΔT0.147 = 1 \times 1 \times \Delta T0.147=1×1×ΔT ΔT=0.147∘C\Delta T = 0.147^\circ \text{C}ΔT=0.147∘C

  1. Check options
  • A: 0.147∘C0.147^\circ \text{C}0.147∘C ✅
  • B: 14.76∘C14.76^\circ \text{C}14.76∘C ❌
  • C: 1.476∘C1.476^\circ \text{C}1.476∘C ❌
  • D: 0.014∘C0.014^\circ \text{C}0.014∘C ❌

Therefore, the correct option is A.

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