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B) The thermostat of the kettle jams and the water continues to boil.After 800 s only...

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lak-86 | Student, Undergraduate | Salutatorian

Posted July 1, 2013 at 11:06 AM via web

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B)

The thermostat of the kettle jams and the water continues to boil.After 800 s only one-half of the water remains, the rest having turned to vapour. Calculate a value for the specific latent heat of vaporisation of water at 100°C.

Refer part A) also

http://www.enotes.com/homework-help/an-electric-kettle-filled-with-1-50-kg-water-20c-441774

 

 

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted July 1, 2013 at 11:17 AM (Answer #1)

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After 800s only half of the water retains in the kettle. This means inside this 800s the other half of water has vaporized by the heat of the kettle. At vaporization the water absorb latent heat from the kettle.

Heat absorbed by water `= 2.1xx10^3xx800 = 1680KJ`

Amount of water vaporized `= 1.5xx1/2 = 0.75kg`

 

For latent heat;

`1680 = 0.75xxL`

`L = 1680/0.75`

`L = 2240(KJ)/(kg)`

 

So the latetnheat of water is 2240KJ/kg

 

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