An electric kettle takes $4\, A$ current at $220\,V.$ How much time will it take to boil $1\,\, kg$ of water circuits are consequences of from temperature $20\,^o C\, ?$ The temperature of $(a)$ conservation of energy and electric charge boiling water is $100\,^o C.$ ............... $min$
AIPMT 2008, Medium
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$\text { Power }=220\, \mathrm{V} \times 4\, \mathrm{A}=880\, \text { watts. }=880\, \mathrm{J} / \mathrm{s}$
Heat needed to raise the temperature of $1\,kg$ water through $80\,^{o} \mathrm{C}$
$ = ms.\,\Delta T \times 4.2\,{\rm{J}}/{\rm{cal}}$
$ = 1000\,{\rm{g}} \times 1\,{\rm{cal}}/{\rm{g}} \times 80 \times 4.2\,{\rm{J}}/{\rm{cal}}$
$ \therefore $ Time taken ${=\frac{1000 \times 1 \times 80 \times 4.2}{880}=\frac{336 \times 10^{3}}{880}} $
${=382\, \mathrm{s}=6.3\, \mathrm{min}}$
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