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...
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 ?
Ok I think what I dont 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 thats 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...
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...
Homework Statement
find the final temperature of A 150g (0,15kg) cube of ice at 0.0 degrees Celsius is added to 300g(0.3 kg) of water at 50.0 degrees Celsius.
Specific heat capacity of ice: 2.09 x 10^3 J/(kg x degrees Celsius)
Specific heat capacity of water: 4.186 x 10^3 J/(kg x...
[SOLVED] Thermodynamic ( quanity of heat)
Homework Statement
Calculate he quantity of heat that needed to transfer 2,0 kg ice of -10C to steam/vapor of 100C
The process in this case take place in a good heat-isolated container with heat capacity of 500J/Kg
Homework Equations
Q=mcT
Q+-ml...
I dont see why I cant:confused:
And I dont know about that but I need to get the task done so I need to do the way I can :wink: But I would like to see your ways of doing it caus its probably a better and "safer" way.
Found another (easy) way to solve it :)
thin end 1/5* 10= 2
thick end 4/5*10= 8
So the gravity of force is 2m from the thick end and 8m from the thin end.
So now you can reveal your way that Iv tryed to understand for 4 days now :)
I dont know how to explain I dont think I understand some basics here.
why is the 1 one 0 + wX and the second one 10 - x ?
Is there another way to solve it ? =s
http://www2.ucdsb.on.ca/tiss/stretton/PhysicsFilm/ForceTorque/sld011.html
Found this. Could this problem be solved in a...
"The resultant force agains the bottom of the tub" I would say it also have info about a hatch but I think that is only relevant to the next question who is what the resultant force is agains the hatch..
Dont ask me You are the smart one:wink:
But the resultant force to the bottom of the "tub"
I dont know but we got the information about lenght and width and so far i havent used that info, maybe i should :confused: