Question on: JAMB Physics - 2018
An electric heating coil rated at 1Kw is used to heat 2kg of water for 2 minutes. The initial water temperature is 30\(^o\)C. Taking the specific heat of the water as 4,000Jkg \(^{-1}\) and neglecting that of the container, the final water temperature is
Power, P = 1kw=100W
 Time, t = 2mins = 2 × 60 = 120secs
 Energy = P×t
 1000 × 120 = 120,000J
 Energy = mc(θ\(_2\) – θ\(_1\))
 120,000 = 2 × 4000(θ\(_2\) – θ\(_1\))
 θ2 = 30 + 15 = 45 \(^o\) C
Add your answer
Please share this, thanks!
No responses