What is the final temperature of water when ice is added to it?

• SlickNick0013
In summary, the final temperature of the water after adding 10kg of ice at -3 degrees C to a tub of 300kg water at 70 degrees C is approximately 68-69 degrees C. The equation used to solve this problem is Q=CM(Tfinal-Tinitial) for both ice and water, with heat capacity values of 4.184J/cal for water and 2.092J/cal for ice. The error in the original calculation was due to not taking into account the magnitude of the difference between lost and gained heat.
SlickNick0013
1. You have a tub of water that you are placing ice into. There is 300kg of water at 70 degrees celsius. You add 10kg of ice at -3 degrees C. What is the final temperature of the water? The latent heat of fusion for water is 33.5 x 10^4 j/kg

I have worked out the best I can the problem as I understand it and I come up with an answer that does not quite make sense

2.
Q= heat
Qgained = Qlost;
therefore
Qgained - Qlost = 0

3. ice is gaining ...water is losing

mc(changeintemp) + mLfusion - mc(ChangeinTemp) = 0

mc(start - final) + mLfusion - mc(start - final)=0

mc(-3 - final) + mLfusion - mc(70-final)=0

((10*4186(-3-final)) + 33.5x10^4) - (300*4186(70-final))=0

(41,860(-3-final) + 33.5 *10^4) - (1,255,800(70-final))=0

-125,580 - 41,860final +33.5*10^4 - (87,906,000 - 1,255,800final)=0

-125,580 - 41,860final + 335,000 - 87,906,000 + 1,255,800final = 0

-41,860final + 1,255,800final + 209,420 - 87,906,000= 0

-41,860final + 1,255,800final -87,696,580 = 0

-41,860final + 1,255,800final = 87,696,580

final(-41,860 + 1,255,800) = 87,696,580

final(1,213,940) = 87,696,580

final = 87,696,580 / 1,213,940 = 72.241 degrees C??

if the -41,860 was positive it came to a plausable answer. Now, I've checked over and and over and I don't understand where my error is but... 87,696,580/ (41,860 + 1,255,800) = 67.5 degrees C. Is one of my signs backwards or did i setup the problem incorrectly. Thanks for any help.

Last edited:
/bump

Halp...I'll <3 u long time

mc(-3 - final) + mLfusion - mc(70-final)=0

Seems like it should be mc(-3 + final).

This is the heat gained by the ice. The final temperature will be positive, so as you have it now, heat gained by ice comes out negative. Change that minus to a plus, and heat gained by ice becomes positive, like it should be.

This is an alternative way to do it I suppose: I can't promise that it will work, but it might give you the correct answer. I am not an expert or anything but my last chapter was dealing with Thermodynamics and I had a lot of problems like this.

Q = CMT
Heat = heat capacity * mass * (Tfinal - Tinitial)
Waters heat capacity is 4.184 and ice is 2.092

4.184J/cal * 300kg (x-70C) = 2.092J/cal * 10kg (x-(-3))

Good luck, and also I agree with turning it positive.

Edit: I ran my calcs and I came up with around 68-69C.

Last edited:
I understand the error now... in the equation Qlost = Qgained it isn't actually Qlost = Qgained but rather the magnitude of difference of Qlost and Qgained is equal I.E.

|Qlost| = |Qgained| which sounds dumb because obviously one of them is going to be negative(losing heat) and one positive(gaining heat) but it just never clicked.

So, in bringing one of them to the other side I needed to set up so they both ended up positive (or take their abs value). So for lost it's (InitalT - finalT) but for gained it's (finalT - initialT). I knew there was an error but I just could not understand why in actuallity it is what it is...now I do, thanks!

What is heat transfer?

Heat transfer is the movement of thermal energy from one object to another due to a temperature difference. It can occur through three main mechanisms: conduction, convection, and radiation.

How does heat transfer occur in ice and water?

In the case of ice in water, heat transfer occurs through conduction and convection. The water molecules, which are at a higher temperature, transfer their thermal energy to the ice molecules, causing them to gain energy and melt. The melted ice then mixes with the water through convection, further distributing the heat energy.

What factors affect the rate of heat transfer in ice and water?

The rate of heat transfer in ice and water is affected by several factors, including the temperature difference between the ice and water, the surface area of the ice, the thermal conductivity of the materials, and the presence of any insulating layers.

Why does ice float on water?

Ice floats on water because it is less dense than liquid water. When water freezes, its molecules form a crystalline structure that is less tightly packed than the molecules in liquid water. This results in ice having a lower density and therefore floating on top of the more dense water.

Can heat transfer be reversed in ice and water?

Yes, heat transfer can be reversed in ice and water. If the temperature of the water is lowered below the freezing point of water, the heat energy will transfer from the water to the ice, causing it to freeze again. This process is known as freezing or solidification.

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