CGPET 2017 Solved Paper

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=

V2

R

t
21000=

(220)2

484

t
t=

21000×484

2202

=210s