Question

How much heat is evolved in converting 1.00 mol of steam at 155.0 ∘c to ice at -50.0 ∘c? the heat capacity of steam is 2.01 j/(g⋅∘c) and of ice is 2.09 j/(g⋅∘c)?

Answer 1

When the specific heat capacity of the water is 4.18 J/g.°C so, we are going to use this formula to get the heat for cooling three  phases changes from steam to liquid and from liquid to ice (solid) :

when Q = M*C*ΔT 

Q is the heat in J

and M is the mass in gram = 1 mol H2O * 18 g/mol(molar mass) = 18 g

C is the specific heat J/g.°C

ΔT is the change in temperature

Q = Mw *[ ( Csteam * ΔTsteam)+(Cw*ΔTw) + (Cice  * ΔT ice)]

    = 18 g * [(2.01 * (155-100°C)) + (4.18 * (100-0°C)) + (2.09 * (0 - 55 °C))]

∴Q = 7444.8 J

and when we know that the heat of fusion for water = 334J/g

and heat of vaporization for water =  2260J/g


∴Q for the two phases changes = M * (2260+334) 

                                                      = 18 * (2260+334)

                                                      = 46692 J 

∴ Q total = 7444.8 + 46692 = 54136.8 J