Question

Calculate the amount of energy in kilojoules needed to change 459 g of water ice at −10 ∘C to steam at 125 ∘C. The following constants may be useful: Cm (ice)=36.57 J/(mol⋅∘C) Cm (water)=75.40 J/(mol⋅∘C) Cm (steam)=36.04 J/(mol⋅∘C) ΔHfus=+6.01 kJ/mol ΔHvap=+40.67 kJ/mol Express your answer with the appropriate units. View Available Hint(s)

Answer 1

Answer: $1.41\times 10^3kJ$

Explanation:-

The conversions involved in this process are :

$(1):H_2O(s)(-10^0C)\rightarrow H_2O(s)(0^0C)\\\\(2):H_2O(s)(0^0C)\rightarrow H_2O(l)(0^0C)\\\\(3):H_2O(l)(0^0C)\rightarrow H_2O(l)(100^0C)\\\\(4):H_2O(l)(100^0C)\rightarrow H_2O(g)(100^0C)\\\\(5):H_2O(g)(100^0C)\rightarrow H_2O(g)(125^0C)$

Now we have to calculate the enthalpy change:

$\Delta H=[n\times c_{ice}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[n\times c_{water}\times (T_{final}-T_{initial})]+n\times \Delta H_{vap}+[n\times c_{steam}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = enthalpy change = ?

m = mass of water = 459 g

$c_{s}$ = specific heat of ice = $36.57J/mol^0C$

$c_{l}$ = specific heat of water =  $75.40J/mol^0C$

$c_{g}$ = specific heat of steam = $36.04J/gmol^0C$

n = number of moles of water = $\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{459g}{18g/mole}=25.5mole$

$\Delta H_{fusion}$ = enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole

$\Delta H_{vap}$ = enthalpy change for vaporization = 40.67 KJ/mole = 40670 J/mole

Now put all the given values in the above expression, we get:

$\Delta H=[25.5mole\times 36.57J/mol^0C\times (0-(-10))^0C]+25.5mole\times 6010J/mole+[25.5mole\times 75.40J/mol^0C\times (100-0)^0C]+25.5mole\times 40670J/mole+[25.5mole\times 36.04J/gmol^0C\times (125-100)^0c]$

$\Delta H=1414910.85J=1.41\times 10^3kJ$    

(1kJ = 1000 J)

Therefore, the enthalpy change is $1.41\times 10^3kJ$