Question

if 10 g ice at -10 degrees celsius is placed in 200 g of water at 80 degrees celsius in an insulated container, what will be the temperature of the system when equilibrium is established? (sp. heat H2O(s)= 2.09 /g C , Sp. H2O(l)=4.18J/g C , heat of fusion H2O(s)= 333 J/g)

Ans: 72 degrees Celsius.

Would anyone be able to show me how to get this answer?? This is a level three problem and I've tried it out and I'm not getting the answer. Can anyone shed some light?? Please and thank you!

Answer 1

there is an awesome animation (along with equations) here http://www.hk-phy.org/contextual/heat/cha/la_he/mixing_e.html

if you still need help, post again

Answer 2

Essentially, you need to find the energy it takes to:

1. get the -10 C ice to 0 C ice (q=mCdT)

2. get the 0 C ice to melt (q=m*Hf)

Add these 2, the q resulting from it will be the loss of heat of the 80 C water, you will find the final temperature (q=mCdT)

Now all you need to do is find the final temp of mixing the 10 g of 0 C water and the 200 g of water at the temperature you previously found:

q=mCdT, equate both systems:

cold water hot water
mC(Tf-Ti)=mC(Tf-Ti)