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 : The correct option is, (a)

Solution :

The conversions involved in this process are :

Now we have to calculate the enthalpy change.

where,

= enthalpy change or heat required = ?

m = mass of water = 5.00 g

= specific heat of liquid water =

= specific heat of liquid water =

n = number of moles of water =

= enthalpy change for vaporization = 40.67 KJ/mole = 40670 J/mole

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

Therefore, the enthalpy change is,

"hydrogen bonding is the attraction of h atoms in the water molecule (slight + charge) to o atoms in a neighboring water molecule (slight - charge); it is a weak interatomic force. but it is still strong enough to the water molecules "stick" together."