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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

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

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

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

ElectricPavlov  Dec 10, 2018

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