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JEE Main 201912 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
12 April
Shift
Evening Shift
Subject
Physics
Chapter
Thermal Properties of Matter

JEE Main 2019 · 12 April · Evening Shift · Physics · Thermal Properties of Matter

1kg of water, at 20 ° C is heated in an electric kettle whose...

1kg1kg of water, at 20°C20 ° C is heated in an electric kettle whose heating element has a mean (temperature averaged) resistance of 20Ω.20Ω . The rms voltage in the mains is 200V200V. Ignoring heat loss from the kettle, time taken for water to evaporate fully is close to
[ Specific heat of water =4200Jkg−1°C−1= 4200J{kg}^{- 1} ° {C}^{- 1} Latent heat of water =2260kJkg−1= 2260kJk{g}^{- 1} ]

Options

  1. A3min3min
  2. B16min16min
  3. C22min22min
  4. D10min10min

Correct answer

C. 22min22min

Step-by-step solution

Given specific heat of the water 4200Jkg−1°C−14200J{kg}^{- 1} ° {C}^{- 1}

Latent heat of the water =2260kJkg−1= 2260kJk{g}^{- 1}

Q=P×tQ = P \times t

Q=mc\DeltaT+mLQ = mc\DeltaT + mL

⇒P×t=mc\DeltaT+mL⇒ P \times t = mc\DeltaT + mL

⇒Vrms2R×t=mc\DeltaT+mL⇒ \dfrac{{V}_{rms}^{2}}{R} \times t = mc\DeltaT + mL

⇒(200)220×t=4200×80+2260×103⇒ \dfrac{{\left(200\right)}^{2}}{20} \times t = 4200 \times 80 + 2260 \times {10}^{3}

⇒t=1298s≃22min⇒ t = 1298\text{s} ≃ 22min
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