Mixing Steam & Ice: Solving Qgain=-Qlost

[tanzerino]

[PLAIN]http://img706.imageshack.us/img706/981/piccj.png
I tried writing the equation Qgain=-Qlost but i don't know how to deal with steam it doesn't have a (c)

 

[Cilabitaon]

I tried writing the equation Qgain=-Qlost but i don't know how to deal with steam it doesn't have a (c)[/QUOTE]

I'm not sure what you mean by "doesn't have a (c)"; are you referring to specific heat capacity?

perhaps you could use Q=ml for the steam; as it will begin to cool and form water, and the question states 'the heat of fusion of ice is 3.33e5 J/kg'.

Remeber for the specific heat capacity of water to use grams instead of kilograms (it's the Chemist's fault!)

 

[mgb_phys]

You have the correct approach, the ice takes in a certain energy to melt the steam gives out a certain energy when it condenses.
Then you have hot and cold water to simply find the resulting temperature of.
(since the ice is at 0 and the steam at 100 - you don't need to worry about c for ice/steam since they don't change temperature)

 

[tanzerino]

i don't understand how i can not worry abt the c for ice/steam how woul the qgain=-qlost look like without c for steam. mass of steam multiplied change in temp only :s

 

[tanzerino]

25*3.33e5+25*4186*100+25*2.26e6=-(5*(tf-100))

 

[mgb_phys]

melt 25kg of ice uses= 25*3.33e5 J/kg = 8.325 MJ
condense 5kg of steam gives 5*2.26E6J/kg = 11.3MJ

So we know we have enough energy to melt the ice and we will end up with liquid water!
Now we just have a simple energy start = energy at end question.
To simplify we will work from 0degC, so we say that water at 0deg has no energy

Before (11.3 - 8.325)MJ + 5kg*100C*4186J/kg/C = 5.06MJ
and this goes to heating the 30kg of water from 0C

 

[tanzerino]

ok so i did some wide shots but i donno.i tried this the specific heat for water vapour at constant volume is 31.4 for each mole and the 5 kg of steam has 18*5000 moles so the quation becomes:
25*3.33e5+25*4186*100+25*2.26e6=-(5000*18*31.4(tf-100)) which gives temprature of 73 celsius close but not an answer:(

 

[tanzerino]

so your way gives 40.3 which is approximately one of the answers

 

[tanzerino]

i didnt quite understand the last step though

 

[tanzerino]

Before (11.3 - 8.325)MJ + 5kg*100C*4186J/kg/C = 5.06MJ

 

[tanzerino]

so Before (11.3 - 8.325)MJ is equalizing the whole system into water and this is the excess energy used in heating water what is the 5kg*100C*4186J/kg/C = 5.06MJ

 

[tanzerino]

Any idea?

 

[tanzerino]

i get it now so we balanced them all into one 30kg mixture at 0 celsius and check the excess energy that will heat the mixture.the melting of ice consumed some heat but condensation of steam or the energy inside steam was much greater which provided extra energy but after condensation the steam is a 100 so it will still give out energy to become 0 c and calculating all this extra energy and use it to heat the water at 0c.Thanks a lot ,though not sure if it is definetaly correct cause kinda complicated any say on that?.