How long will it take to heat 3kg of water from 28oC to 88oC in an electric kettle taking 6a from a 220V supply? [specific heat capacity of water = 4180 J kg-1K-1]
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Correct Answer: Option B
Explanation:
t = \(\frac{mc\theta}{IV} = \frac{3 \times 4189 \times 60}{6 \times 120} = 570s\)
t = \(\frac{mc\theta}{IV} = \frac{3 \times 4189 \times 60}{6 \times 120} = 570s\)