An electric kettle takes 4 A current at 220 | vedclass

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(C) Given: Current $I = 4 \ A$, Voltage $V = 220 \ V$, Mass of water $m = 1 \ kg$, Initial temperature $T_1 = 34^{\circ} C$, Final temperature $T_2 =

Vedclass · Accepted Answer

$5.25$

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(C) Given: Current $I = 4 \ A$, Voltage $V = 220 \ V$, Mass of water $m = 1 \ kg$, Initial temperature $T_1 = 34^{\circ} C$, Final temperature $T_2 = 100^{\circ} C$.Specific heat capacity of water $c = 4200 \ J/(kg \cdot ^{\circ} C)$.Electric power $P = V 	imes I = 220 	imes 4 = 880 \ W$.Heat energy required $Q = mc\Delta T = 1 	imes 4200 	imes (100 - 34) = 4200 	imes 66 = 277200 \ J$.Since electric energy is converted into heat energy, $P 	imes t = Q$.$880 	imes t = 277200$.$t = 277200 / 880 = 315 \ seconds$.To convert time into minutes: $t = 315 / 60 = 5.25 \ minutes$.

An electric kettle takes $4 \ A$ current at $220 \ V$. If the entire electric energy is converted into heat energy, then the time (in minutes) taken to increase the temperature of $1 \ kg$ of water from $34^{\circ} C$ to $100^{\circ} C$ is

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