Calculate the time taken to heat 2kg of water from 50°C to 100°C in an electric kettle taking 5A, from a 210V supply. (Specific heat capacity of water = 4200\(kg^{-1}K{-1}\)
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Correct Answer: Option C
Explanation:
t = \(\frac{mc\theta}{p}= \frac{2\times4200\times50}{210\times5} \\ = 400s = 6.7\)
t = \(\frac{mc\theta}{p}= \frac{2\times4200\times50}{210\times5} \\ = 400s = 6.7\)