**What is the equlibrium temperature when an ice cube at -5º C of mass 10g is placed into 200 grams of water at 45º C?**

I need help finding what equatons I need to use to solve this problem or perhaps the answer.

Cent0rea88 Dec 9, 2018

#1**+2 **

Well....that 'ain't' much ice......it will warm up and then melt at constant temp and cool the entire mass (200 + 10gm) of a bit

You'll need Specific heat of ice (0.5 cal /gm-C or 2093 j/kg-c )

10 gm = .01 kg ice

.01 kg (2093 j/kg-c)(5 C) = 104.65 j to warm the ice to 0 DEGREES

Now you'll need the latent heat of fusion of ice to melt it .336 MJ/kg

.01kg (.336 Mj/kg) = 3360 j

The heat to warm the ice and melt it has to come from the water

the final mass of water ( 200+10) will cool off an amount in joules equal to the two amounts we just calc'd

So, you'll need Specific heat of water 4.186 j/gm c

210gm (4.186 j/gm-c) (45-x c ) = 104.65 + 3360 j

45-x = 3.9413 (the water loses 3.9413 degrees)

41.059 degrees C is equalibrium

Corected....changed 220 to 210 in last calcs.....

ElectricPavlov Dec 9, 2018

#2**+1 **

Thank you so much! I'm pretty good at understanding chemistry, but I can't do math for the life of me.

Edit: In your answer you listed the total mass as 220 grams instead of 210. Is that correct?

Cent0rea88
Dec 10, 2018

#3**+2 **

Sorry....that should have been 210 gm....I have corrected it......THANX !

ElectricPavlov
Dec 10, 2018