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### thermodynamic

Posted: **Mon Mar 05, 2012 1:50 pm**

by **wrghcharl**

A certain pressure cooker has a volume of 6L and an operating pressureof 75kpa. initally, it contains 1kg of h2o. heat is supplied to the pressure cooker at a rate of 500w for 30 mins after the operating pressure is reached. assuming an atmospheric pressure of 100kpa ateremining (a) the temp at which cooking takes place. (b) the amount of h20 left in the pressure cooker at the end of the process.

### Re: thermodynamic

Posted: **Sat Mar 17, 2012 12:35 am**

by **vikrambijarniya**

wrghcharl wrote:A certain pressure cooker has a volume of 6L and an operating pressureof 75kpa. initally, it contains 1kg of h2o. heat is supplied to the pressure cooker at a rate of 500w for 30 mins after the operating pressure is reached. assuming an atmospheric pressure of 100kpa ateremining (a) the temp at which cooking takes place. (b) the amount of h20 left in the pressure cooker at the end of the process.

operating pressure of pressure cooker = 75kpa

so pressure inside pressure cooker = 75+100(atmospheric pressure) = 175 Kpa

at this pressure boiling point of water is around 115 degree celcius (which cant be calculated by given data)

now total heat supplied after operating pressure is reached is = power * time = 500*30*60 = 900 kilojoules

latent heat of vaporization (L) = 540 cal/gram = 540*4.2= 2268 joules/gram = 2268KJ/kg

so 2268KJ of heat will vaporize 1kg of water

then 1KJ of heat will vaporize = 1/2268 kg of water = 0.000441 kg of water

then 900KJ of heat will vaporize = 900*0.000441 kg = .397 kg ~ .4kg = 400grams

so amount of water left = 1000-400 = 600grams

### Re: thermodynamic

Posted: **Sat Mar 17, 2012 4:38 am**

by **giga17**

Like seriously, do your own homework. How are you going to learn otherwise?