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.
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.....