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`        An immersion heater rated 1000 W, 220 V is used to heat 0.01 m^3 of water. Assuming that the power is supplied at 220 V and 60% of the power supplied is used to heat the water, how long will it take to increase the temperature of the water from 15°C to 40°C?`
3 years ago

396 Points
```										Sol. P = 1000 V = 220 v R = V^2/P = 48400/1000 = 48.4 Ω
Mass of water = 1/100 * 1000 = 10 kg
Heat required to raise the temp. of given amount of water = ms∆t = 10 * 4200 * 25 = 1050000
Now heat liberated is only 60%. So V^2/R =T=60% = 1050000
⇒ (220)^2/48.4 * 60/100 * T = 1050000  ⇒ T = 10500/6 * 1/60 nub = 29.16 min.

```
3 years ago
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