You balance the heat equation.
heat gained by melting ice+heat gained by remaining ice warming up + heat gained by hot water=0
This of course means that the nonmelted ice is at a final temperature of zero, otherwise there cannot be an ice/water mxture.
Heat gained by melting ice:
(70-m)Hf+ (70-m)cice(8)
Heat gained by nonmelting ice:
(m*cice(8)
heat gained by hot water: 100*cwater(0-20)
add them, set to zero, solve for m.
Suppose 100 g of water at 20°C is poured over a 70-g cube of ice with a temperature of -8°C. If all the ice melts, what is the final temperature of the water? If all of the ice does not melt, how much ice remains when the water–ice mixture reaches equilibrium?
I'm aware that not of the ice melts, but how do I determine how much ice is left?
6 answers
I'm getting -25.1 g when I use this equation.
.07*334kJ-m334+.07*2.1*8+.1*4.18*(-20)=0
m= - 1/334 * (.418*20-.07*334-.07*16.8)
which is not a negative number....
check my work, most of it was in my head.
m= - 1/334 * (.418*20-.07*334-.07*16.8)
which is not a negative number....
check my work, most of it was in my head.
I forgot that 70g was 0.07 kg. I was using 0.007 kg. Thank you. This question makes a lot more sense now.
7.896 degrees celsius
7.896 degrees celsius and 15.125 g of ice