Answer:
Energy consumed by the electric kettle in 9.5 min =Pt=(2.5×10
3
)×(9.5×60)=14.25×10
5
J
Energy usefully consumed =msΔT=3×(4.2×10
3
)×(100−15)=10.71×10
5
where s=4.2J/g
o
C= specific heat of water and boiling point temp=100
o
C
Heat lost =14.25×10
5
−10.71×10
5
=3.54×10
5
