Question

10.0 g of ice at 0.00°C is mixed with 25.0 g of water at 35.00°C in a coffee-cup calorimeter. What is the final temperature of the mixture? The specific heat of water is 4.18 J/g-°C; the heat of fusion of water is 333 J/g?

Answer 1

Step-by-step explanation:

The joules gained by ice and warming the resultant water are

10 * 333 + 10 (4.18 )(T-0) Where T is the final temp above 0 C

Joules given up by 25 g of water at 35 C being cooled to T

25 * 4.18 * ( 35- T)

These quantities have to be equal :

10* 333 + 10 (4.18) T = 25 * 4.18 ( 35-T)

solve for T = 2.24 degrees C