JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Current Electricity

One kg of water, at

2 0^{\circ} C

, is heated in an electric kettle whose heating element has a mean (temperature averaged) resistance of

20Ω

. 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 / kg^{\circ} C

), Latent heat of water

= 2260 kJ / kg

]

(A)3 minutes
(B)10 minutes
(C)22 minutes✓ Correct
(D)16 minutes

Answer: (C)

Solution

The element dissipates

P = \frac{V ^{2}}{R} = \frac{20 0 ^{2}}{20} = 2000

W. Raising the water from

2 0^{\circ}

C to

10 0^{\circ}

C needs

m c Δ T = 1 \times 4200 \times 80 = 3.36 \times 1 0^{5}

J, and boiling it all away needs

m L = 1 \times 2.26 \times 1 0^{6} = 2.26 \times 1 0^{6}

J. The total

2.596 \times 1 0^{6}

J delivered at

2000

W takes

\frac{2.596 \times 1 0 ^{6}}{2000} \approx 1298

s, close to

22

minutes.

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Current Electricity

Solution

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