The ice acquires 50 g * 80 Cal/g = 4000 Cal from the water while melting, but remining 0 C. In losing that amount of heat, the original liquid water cools by 4000 Cal/(200g*1 Cal/deg C) = 20 C, leviong it at 20 C.
Now you mix 200 g of water at 20 C with 50 g at 0 C.
At equilibrium T,
50*T = 200*(20 - T)
250 T = 4000
T = 16 C
you chop 50 g of ice at 0 degrees celsius into 200g of water at 40 degress celcius. What is the final temperature?
1 answer
1 answer