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There is an electric kettle with the specification of "220V 1000W". When the electric kettle works normally, what is the resistance of the internal heating wire? 1. How long does it take to boil 1kg water at 20 ℃ under rated voltage? (excluding heat loss) 2. If the actual heating time is 6min, what is the thermal efficiency of the electric kettle?
1. Resistance of heating wire: r = 220 ^ 2 / 1000 = 48.4 ohm
2. Q suction = cm Δ t = 4.2 × 10 ^ 3 × 1 × (100-20) = 3.36 × 10 ^ 5J
3. Electric energy consumed by electric heater: w = Pt = 1000 × 60 × 6 = 3.6 × 10 ^ 5J
Therefore, the thermal efficiency of the electric kettle is: η = q absorption / w = 3.36 × 10 ^ 5 / 3.6 × 10 ^ 5 = 93.3%
An electric kettle marked with 220V 1000W, if it works normally for 7min and all the heat released is absorbed by water, how much ℃ can the temperature of 2kg water rise from 20 ℃ (C water 4.2 * 10 ^ 3j / (kg '℃))
Pt=cM(t-20)
1000W*7*60s=4.2*10^3J/(kg`℃）*2kg*(t-20)
The solution is t = 70 ℃
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