Question

How many grams of ice at -9.7 ?C can be completely converted to liquid at 24.2 ?C if the available heat for this process is 4.95×10³ kJ ?

For ice, use a specific heat of 2.01 J/(g??C) and ?Hfus=6.01kJ/mol.

Answer 1

The conversion of ice at - 9.7 degreeC to liquid at 24.2 degreeC takes place in the following way .

(a) Firstly ice at -9.7 degreeC gets heated to ice at 0 degreeC

Q1 = m * c * deltaT

Q1 = m * 2.01 J / g-degreeC * [ 0 - ( - 9.7) ] degreeC

Q1 = 19.5 * m   J

(b) Ice is then converted into liquid at 0 degreeC

Q2 = m * [ delta H(fusion) / molar mass of water ]

Q2 = m * (6.01 x 10^3 J/mol / 18 g/mol)

Q2 = m * 333.89 J

(c) Water at 0 degreeC is then converted to water at 24.2 degreeC

Q3 = m * 4.18 J/g-degreeC * ( 24.2 - 0 ) degreeC

Q3 = m * 101.156 J

Now ,

Total heat required = Q1 + Q2 + Q3

4.95 x 10^3 kJ = 4.95 x10^6 J

4.95 x10^6 J = m ( 19.5 + 333.89 + 101.156)

m = 1.089 x 10^4 g = 1.09 x 10^4 g ( in three significant figures)

So, mass of ice required for this process = 1.09 x 10^4 g