Recent content by Newton86

urgent. Growth in salmon smolt Homework Statement smolt weith is 35g we shall find the weight after 61 days. Homework Equations specific growth rate:; K' = Ln (N2 / N1) / (t2 - t1) growth rate is 1,9 % The Attempt at a Solution I find 110g when I use this forumula...

Yeah actually :) Only I did not put them up that way

why nobody ?

any?

This should be a easy task for you :p

Help :)

:D so that's the answer then 3859000J Thanks for the help :)

Ok :) thanks for the advice. But what do I now Is it just to add the heat from the container 55000J too the 4 parts added = 3804000J = 3804000J + 55000J = 3859000J ?

hmm Starting temprature of the container must be -10 Celsius and ending temprature of 100Celsius ? So then the heat from container is 500j/k *110K = 55000J ?

did I do something right here?

Ok I think what I don't seems to find now is the Also set your equations equal to one another. Energy lost = energy gained ect...I had the same problem awhile back and that's what I was leaving out. hmm is the energy lost in the phase changes ?

What I tryed is: Water Q=mt = 0,3*4190 = 1257J Ice to liquid: Q=mt = 0,15*334000 = 50100J 1) The melted Ice is now water of 0C = (0,15kg)*(4190)j/kg*K)(T-0) 2) The water of 50c = (0,3kg)*(4190)j/kg*K)(T-50) solving that (0,15)*(4190)= 628,6T...

hmm. So what I do is to find the temprature out of thoose 4 parts and * it with the 500J/K ?

Aii yes it is 500J/K not Kg :) I tryed to find the heat gained from ice to vapor. 1: Ice heats up to melting point Q=mcdeltaT = 2kg*2100j/kg*K*10K = 42000J 2: Ice melts at C0 Q=+-ml = 2kg* 334000J = 668000J 3: liquid water of C0 heats up to 100C Q=mcdeltaT = 2kg*4190j/kg*K*100K...

Hmm =| Off course the container is inportnant. I mixed up. It has a heat capacity of 500J/Kg But it doesn't have a mass. So what I do with it ?