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* 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* c_(p,l)* (T_(final)-T_(initial))]+n* \Delta H_(vap)+[m* c_(p,g)* (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}}=(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* 4.18J/g^oC* (100-14)^oC]+0.278mole* 40670J/mole+[5.00g* 1.84J/g^oC* (115-100)^oC]`

`\Delta H=13241.66J=1.32* 10^4J`

Therefore, the enthalpy change is,
`1.32* 10^4J`