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` ?

Answer

B

Explanation

`P = 1000W, V = 220V`
`R = (48400)/(1000)`
` `
`= 48.4 Omega`
` Mass of water `
`=(1)/(100)xx1000 = 10 Kg`
` Specific heat of water = 4200 Heat required to raise the temp. of given amount of water `
`=mstheta = 10xx4200xx25`
` =105000 Now heat liberated is only 60% So, `
`((V^2)/(r )) xx t xx 60% = 1050000`
` `
`rArr((220))^2/(48). (txx60)/(100) = 1050000 `
`rArr t = 29 minutes`.