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# I need help with - Current Electricity - JEE Main

One kg of water, at $20^{\circ}C$, is heated in an electric kettle whose heating element has a mean (temperature averaged) resistance of $20\Omega$. 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 =$4200J/(kg^{\circ}C)$, Latent heat of water = 2260 kJ/kg]

• Option 1)

16 minutes

• Option 2)

22 minutes

• Option 3)

3 minutes

• Option 4)

10 minutes

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Heat developed in a resistor -

When a steady current flows through a resistance R for time t , the loss in electric potential energy appears as increased thermal energy(Heat H) of resistor and $H=i^2Rt$

The power devoleped = $\frac{energy}{time}=i^2R=iR=\frac{V^2}{R}$      (from Ohm's law)

- wherein

Unit of heat is joule (J)

Unit of power is watt (W)

$ms\Delta T+mL=\frac{V^{2}}{R}t$

$1\times 4200\times (100^{\circ}-20^{\circ})c+1\times 22.6\times 10^{5}=\frac{(200)^{2}}{20}t$

$336\times 10^{3}+226\times 10^{3}=2\times 10^{3}t$

$t=1298 sec. = 21.6 min \approx 22 minutes$

Option 1)

16 minutes

Option 2)

22 minutes

Option 3)

3 minutes

Option 4)

10 minutes

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