Question

How many kilojoules are required to warm 25.0 g of water from 75.0 C to 100.0 C and convert it to steam at 100.0 C? The heat of vaporization of water is 2260 joules/g, and the specific heat of water is 4.184 joules/g.

Answer 1

Heat of Vaporization can be defined as the total amount of heat required to convert the liquid state into vapor without increasing the temperature of that liquid or fluid i.e. that heat which changes the liquid state into gaseous state.

Complete answer:
Based on entropy and enthalpy of vaporization and relationship between them heat of vaporization can be calculated by the following formula: $\text{Q=2364}\text{.6}\times \text{25=59115J=59}\text{.1KJ}$
${{H}_{v}}=\dfrac{q}{m}$
Where, ${{H}_{v}}$represents the heat of vaporization, m is mass of substance and q is heat absorbed.
To heat 1.0 g of water from 75.0 C to 100.0 C which specify that temperature rise will be equal to 25.0 C, so we can say that heat required will be equal to
$Rise\ \text{in temp }\times \text{ specific heat = 25 }\times \text{ 4}\text{.184 = 104}\text{.6 J/g}$
Heat of vaporization of water is 2260 joules/g (Given)
So total energy required to heat up and vaporize 1 gram would be equal to the sum of both i.e.
$\text{ 104}\text{.6 +2260 = 2364}\text{.6 J/g}$
The value of heat required i.e. Q can be find by putting the value of ${{H}_{v}}$and m in the above mentioned formula
i.e. $\text{Q=2364}\text{.6}\times \text{25=59115J=59}\text{.1KJ}$, m is 25 g (Given)
Hence 59.1 kilojoules are required to warm 25.0 g of water from 75.0 C to 100.0 C and convert it to steam at 100.0 C.

Note:
Heat of vaporization can also be defined as the quantity of heat that needs to be absorbed to vaporize a particular quantity of liquid at a constant temperature. If the solutions of vapor and liquid states are compared then the kinetic energy of the steam is always comparatively higher than the kinetic energy of the fluid.