An electric kettle marked with "220 V & nbsp; 550 W" works under rated voltage. The calculation is as follows: (1) the current through the electric kettle when it works normally; (2) the electric energy consumed by the electric kettle when it is energized for 6 min

An electric kettle marked with "220 V & nbsp; 550 W" works under rated voltage. The calculation is as follows: (1) the current through the electric kettle when it works normally; (2) the electric energy consumed by the electric kettle when it is energized for 6 min

(1) ∵ u = 220 V, P = 550 W, ∵ according to the formula P = UI, the current of the electric kettle in normal operation is: I = Pu = 550 W, 220 V = 2.5 A; (2) power on T = 6 min = 360 s, the electric energy consumed by the electric kettle is: w = Pt = 550 w × 360 s = 1.98 × 105 J. (1) the current of the electric kettle in normal operation is 2.5 A; (2) the electric energy consumed by the electric kettle in normal operation for 6 min is 1.98 × 105 J

The electric kettle has a capacity of 2l, a rated power of 800W, and a rated voltage of 220V. Q: (1) how much heat does it generate after 420s of normal operation? (2) the electric kettle is full
How many degrees Celsius can the heat generated in (1) increase the water?
(3) If the electric kettle has heat preservation function and the heat preservation power is 50W, what resistance should be connected in series?

V = 2 L = 2 * 10 ^ (- 3) m3, P = 800 W, u = 220 V, t = 420 s, C water = 4.2 * 10 ^ 3 J / (kg ℃), ρ water = 1 * 10 ^ 3 kg / m3. Analysis: (1) the heat generated by electric kettle in 420 s is q = P * t = 800 * 420 = 3.36 * 10 ^ 5 J

An electric kettle marked with "220V 1.1KW" works under rated voltage. How much heat does the electric kettle produce when it is electrified for 10min?

It can be seen from the formula: q = IRT = uit = Pt

For an electric kettle marked "220 V 1.1 kW", calculate:; resistance of the kettle; heat generated by the electric kettle at the same point for 10 minutes

From P = UI and I = u / R, we can get r = u & # 178 / P to calculate the resistance of the kettle: r = u & # 178 / P = (220 V) &# 178 / 1100W = 44 Ω, from P = w / T to get w = Pt, and to calculate the heat generated after 10 minutes of power on: q = w = Pt = 1100W × 600s = 660000j