An electric kettle takes $4 A$ current at $220 V$ . How much time will it take to boil $1 k g$ of water from temperature $20^{\circ} C$ ? The temperature of boiling water is $100^{\circ} C$ .

8.4 m i n

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4.2 m i n

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6.3 m i n

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12.6 m i n

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Solution

The correct option is C $6.3 m i n$
Formula used: $Q = m s Δ θ, Q = V i t$

Given: $i = 4 A, V = 220 V, m = 1 k g, θ_{1} = 20^{\circ} C, θ_{f} = 100^{\circ} C, s = 4200 J / k g^{\circ} C$

Heat energy released by resistor gets used in increasing temperature of body.
Since we already know about how much increase in temperature $Δ θ = 100 - 20 = 80^{\circ} C$ , we have to find time for which kettle must be kept on.
Energy $= V i t = m s Δ θ$
$220 \times 4 \times t = 1 \times 4200 (100 - 20)$
$t = 381.85 s = 6.3 m i n$

FINAL ANSWER: (c)

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An electric kettle takes 4 A current at 220 V. How much time will it take to boil 1 kg of water from temperature 20∘C? The temperature of boiling water is 100∘C.

An electric kettle takes $4 A$ current at $220 V$ . How much time will it take to boil $1 k g$ of water from temperature $20^{\circ} C$ ? The temperature of boiling water is $100^{\circ} C$ .