Answer to Question #100428 in General Chemistry for ghazl

Question

how to calculate enthalpy change for the process in which 50 g of water is converted from liquid at 10°C to vapor at 25°C

Answer 1

Here we want to vaporize water at room temperature, which will require more heat than at $100^∘C$ .

Heat flow is a path function, so we can separate this into specific steps. All of this is done at constant pressure, so the heat flow is equal to the enthalpy, a state function, and the order of these steps doesn't matter.

We set a "boiling point" at $25^∘C$, and thus have two steps:

1. Heat from  $10^∘C$  to  $25^∘C$

2. Vaporize at  $25^∘C$

Step 1 would be a simple heating at constant pressure:

 $q_1=mC_PΔT$

where

 m is mass in  g ,  

$C_P$  is the specific heat capacity at constant atmospheric pressure in  

$J/g^∘C$  

ΔT is the change in temperature until the phase change temperature.

In this case we specify  $T_f=25^∘C$

$\implies q_1 =50*4.184*(25-10)$

$q_1=3138 J=3.138kJ$

Step 2 would be vaporization at constant pressure again, but at a CONSTANT, lower temperature than usual.Recall that  

$q=ΔH$

at constant pressure (provided they have the same units).

Thus, we just have:

$q_2=nΔH_{vap}T$

where,

n is number of moles of water

$\Delta H_{vap}$ is heat of vaporization in $kJ/mol$

T is the temperature at which this phase change occurs.

$\implies q_2=(50/18)*(44)*(25)$

$q_2=3055.555kJ$

$∆H =q_1+q_2$

$∆H=3.138+3055.555$

$∆H=3058.693kJ$ (Answer)