You use a tea kettle (heating coil resistance R = 125 ohm) plugged into the wall

Question

You use a tea kettle (heating coil resistance R = 125 ohm) plugged into the wall (120 V) to bring 210 ml of water from 24 degreeC to boiling temperature 100 degreeC. Compute the electric current in the kettle. Compute the electric power of the kettle. 4190 J of heat ii required to raise the temperature of 1 kg of water 1 degreeC. How much heat is needed to bring 210 ml of water from 24 degreeC to a boil? Compute the time required for the water to go from 24 degreeC to a boil, assuming the kettle's heating coil is 100% efficient. Actually the kettle is only 60% efficient. How much time does it take now?

Explanation / Answer

(a) Current (I) = Voltage (V)/Resistance (R)

= 120/125 = 0.96 A

(b) Power = V2/R

= (120)2/125 = 115.2 115 W

(c) Mass of water = volume x density = 210 mL x 1.00 g/mL

= 210 g = 0.210 kg

Heat required = mass x specific heat x temperature change

= 0.210 x 4190 x (100 - 24) = 66872.4 J 6.69 x 104 J

(d) Power = heat energy/time

115.2 = 66872.4/time

Time = 580 s

(e) Power = heat energy/time

(60/100) x 115.2 = 66872.4/time

Time = 967 s

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