Question

A 10.67-gram -23.0°C piece of ice drops into a calorimeter containing 150.00-gram 93.7 °C of the water. What is the final temperature? (Swater = 4.184 J/g° C, Sice = 2.082 J/g° C, Heat of fusion= 334.7 J/g)

Answer 1

consider all the water at 93.7 C is converted to water at 0 C

we know that

Q = m x s x dT

so

Q = 150 x 4.184 x 93.7

Q = 58806.12 J

now

this heat is used to change the temperature of ice

now

heat required to bring Ice at -23 to O C

Q1 = 10.67 x 2.082 x 23

Q1 = 510.94362 J

now

heat required to melt the ice to water at 0 C

Q2 = m x dHf

Q2 = 10.67 x 334.7

Q2 = 3571.249

so

total heat required to bring Ice at -23 C to water at 0 C

Q3 = Q1 + Q2

Q3 = 510.94362 + 3571.249

Q3 = 4082.19262

unused heat = 58806.12 - 4082.19362 = 54723.92738

now

all the ice is coverted to water and 150 g of water is already present at 0 C

so

total water at 0 C = 150 + 10.67 = 160.67

the usused heat is used to raise the temperature of water

so

Q = m x s x dT

54723.92738 = 160.67 x 4.184 x (T - O)

T = 81.4 C

so

the final temperature of the system is 81.4 C