Question

Calculate the enthalpy change (in joules) involved in converting 5.00 grams of water at 14.0 °C to steam at 115 °C under a constant pressure of 1 atm. The specific heats of ice, liquid water, and steam are, respectively, 2.03, 4.18, 1.84 J/g-K and for water ΔHfusion = 6.01 kJ/mole and ΔHvap = 40.67 kJ/mole
a. 1.32x10^4 J
b. 2.05 x 10^5 J
c. 195x 10^3 J
d. 1.94 x 10^3 J

Answer 1

Answer : The correct option is, (a) $1.32\times 10^4J$

Solution :

The conversions involved in this process are :

$(1):H_2O(l)(14^oC)\rightarrow H_2O(l)(100^oC)\\\\(2):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\\\(3):H_2O(g)(100^oC)\rightarrow H_2O(g)(115^oC)$

Now we have to calculate the enthalpy change.

$\Delta H=[m\times c_{p,l}\times (T_{final}-T_{initial})]+n\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = enthalpy change or heat required = ?

m = mass of water = 5.00 g

$c_{p,l}$ = specific heat of liquid water = $4.18J/g^oC$

$c_{p,g}$ = specific heat of liquid water = $1.84J/g^oC$

n = number of moles of water = $\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{5.00g}{18g/mole}=0.278mole$

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

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

$\Delta H=[5.00g\times 4.18J/g^oC\times (100-14)^oC]+0.278mole\times 40670J/mole+[5.00g\times 1.84J/g^oC\times (115-100)^oC]$

$\Delta H=13241.66J=1.32\times 10^4J$

Therefore, the enthalpy change is, $1.32\times 10^4J$