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JEE Main 2019 (12 Apr, Shift 2) · Calorimetry, Latent Heat & Mechanical Equivalent of Heat

Calorimetry, Latent Heat & Mechanical Equivalent of Heat question from JEE Main 2019 (12 Apr, Shift 2)

Q36JEE Main 2019 · 12 APR · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat

One kg of water, at 20oC, heated in an electric kettle whose heating element has a mean (temperature averaged) resistance of 20 Ω\Omega. The rms voltage in the mains is 200 V. Ignoring heat loss from the kettle, time taken for water to evaporate fully, is close to: [Specific heat of water = 4200 J/(kgoC), Latent heat of water = 2260 kJ/kg]

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Answer: (b)

  1. Power P=V2R=200220=2000 WP = \dfrac{V^2}{R} = \dfrac{200^2}{20} = 2000\text{ W}
  2. Heat =mcΔT+mL=1×4200×80+2.26×106=2.596×106 J= mc\Delta T + mL = 1 \times 4200 \times 80 + 2.26 \times 10^6 = 2.596 \times 10^6\text{ J}
  3. t=2.596×1062000=1298 s≈21.6 mint = \dfrac{2.596 \times 10^6}{2000} = 1298\text{ s} \approx 21.6\text{ min}
  4. So the answer is (b) 22 minutes.
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