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
15.0\(^o\)C
45.0\(^o\)C
52.5\(^o\)C
60.0\(^o\)C
Correct answer is B
Time = 2mins = 2 × 60 = 120secs
Energy = power × time
1000 × 120 = 120,000J
Energy = mc(θ\(_2\) – θ\(_1\))
120,000 = 2 × 4000(θ\(_2\) – θ\(_1\))
θ2 = 30 + 15 = 45 \(^o\) C