let R be resistance of kettle, m be mass of water, c be sp. heat capacity and d@ be change in temp Let the time taken be t In first case, Energy spent=2202/R*5*60 Also, heat energy supplied=mcd@ thus, mcd@=2202/R*5*60.......1 Similarly in second case mcd@=2002/R*t*60.......2 on equating equations 1 and 2, we get t=6.05 minutes or 6 minutes 3 seconds
Moderation Team
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