Question

How much heat is evolved in converting 1.00 mol of steam at 145.0 ℃ to ice at -50.0 ℃? The heat capacity of steam is 2.01 J/(g⋅℃) and of ice is 2.09 J/(g⋅℃).

Answer 1

Cp(steam) = 2.01 J/(g⋅∘C).

Cp(ice) = 2.09 J/(g⋅∘C).

mass of steam = 1 mole steam = 18 g

delta Hfus = 334 J/g

delta Hvap = 2260 J/g

heat Q = m x Cp x dT

1) heat required to cool steam 145 to 100C = 18 x 2.01 x (100 - 145) = -1628.1 J

2) heat of vapourisation at 100 C = 18 x 2260 = 40680 J

3) heat of cooling water to 100 C to 0 C = 18 x 4.184 x (0-100) =7531.2 J

4) heat of melting at 0 C= 18 x 334 = 6012 J

5) heat of cooling ice = 18 x 2.09 x 50 = 1881 J

total heat evolved = Q1 + Q2 + Q3 + Q4 + Q5

                             = 1628.1+40680 +7531+6012+1881

                             = 57732 J

   = 57.7 kJ