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...

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 28^oC to 88^oC: Q = (3 kg) × (4180 J kg^-1K^-1) × (88^oC - 28^oC) 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.