Question

How much heat is absorbed when 63.7 g H2O(l) at 100 degrees Celsius and 101.3kPa is converted to steam at 100

Answer 1

Answer:

Q = 143,921 J = 143.9 kJ.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the absorbed heat by considering this is a process involving sensible heat associated to the vaporization of water, which is isothermic and isobaric; and thus, the heat of vaporization of water, with a value of about 2259.36 J/g, is used as shown below:

$Q=m*\Delta _{vap}H$

Thus, we plug in the mass and the aforementioned heat of vaporization of water to obtain the following:

$Q=63.7g*2259.36J/g\\\\Q=143,921J=143.9kJ$

Regards!