Given Resistance of heating coil =484Ω Supply voltage =220V Mass of water, m=100g=0.1kg Specific heat (s)=4.2J∕g°C Increase in temperature of water (ΔT)=50°C Heat energy required for this raise in temperature is given by O=msΔT=100g×4.2J∕gram°C×50°C =21000J ........(i) Heat dissipated by the heating coil =
V2
R
t =
(220)2
484
t .........(ii) From Eqs. (i) and (ii), we get msΔT=