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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.

 Dec 9, 2018
 #1
avatar+19325 
+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.....cheeky

 Dec 9, 2018
edited by ElectricPavlov  Dec 10, 2018
 #2
avatar+55 
+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
edited by Cent0rea88  Dec 10, 2018
 #3
avatar+19325 
+2

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

ElectricPavlov  Dec 10, 2018

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