How long will it take to heat 3 kg of water from 28^{o}C to 88^{o}C in an electric taking a current of 6 A from an e.m.f. source of 220V?

Answer Details

To solve this problem, we need to use the formula:
Q = m c ΔT
where Q is the heat energy required to heat the water, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
The specific heat capacity of water is 4.18 J/g°C, and since we have 3 kg of water, the mass of water is 3000 g.
So, Q = 3000 g × 4.18 J/g°C × (88°C - 28°C) = 376440 J
Now, we can use the formula for electrical energy:
E = V I t
where E is the electrical energy, V is the voltage, I is the current, and t is the time.
Rearranging the formula to solve for t, we get:
t = E / (V I)
Substituting the given values, we get:
t = 376440 J / (220 V × 6 A) = 570 s
Therefore, the time it will take to heat 3 kg of water from 28°C to 88°C in an electric taking a current of 6 A from an e.m.f. source of 220V is 570 seconds or 9.5 minutes.
So, the correct answer is 570s.