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 k...
Question 1 Report
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]
Answer Details
To solve this problem, we need to use the formula: Q = mcΔT where Q is the amount of heat energy required, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature. First, we need to calculate the amount of heat energy required to heat the water from 28oC to 88oC: Q = (3 kg) × (4180 J kg-1K-1) × (88oC - 28oC) Q = 3 × 4180 × 60 Q = 753240 J Next, we can use the formula: P = VI where P is power, V is voltage, and I is current, to calculate the power of the electric kettle: P = (220 V) × (6 A) P = 1320 W Finally, we can use the formula: t = Q/P where t is time and Q and P are as calculated above, to determine the time taken: t = 753240 J ÷ 1320 W t = 570 s Therefore, the answer is 570s.
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