2g of water at a temperature T1=80 C is mixed with 2g of ice at a temperature of T2=0 C.

I assumed heat lost by the water is heat gained by the ice. -Q(water)= Q(ice)
Where Q(water)= mc∆T
and Q(ice) = mL + mc∆T

I got final temperature is 0 C or 273 K. Anyone back me up?

Question

2g of water at a temperature T1=80 C is mixed with 2g of ice at a temperature of T2=0 C.

I assumed heat lost by the water is heat gained by the ice. -Q(water)= Q(ice)
Where Q(water)= mc∆T 
and Q(ice) = mL + mc∆T

I got final temperature is 0 C or 273 K. Anyone back me up?

australopithecus · Answer

Assuming this is an isolated system wouldn't the final temperature be 40 C, I'm not strong in physical chemistry but it just doesn't make sense to me for the temperature to reach equilibrium at 0 C

anonymous · Answer

Yeah this was 7 months ago! I ended up getting it wrong. I had several of these, but my method worked for some and not others. This was thermodynamics.