How much heat is absorbed when 63.7 g H2O(l) at 100 degrees Celsius and 101.3kPa is converted to steam at 100
0 thoughts on “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:
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:
[tex]Q=m*\Delta _{vap}H[/tex]
Thus, we plug in the mass and the aforementioned heat of vaporization of water to obtain the following:
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:
[tex]Q=m*\Delta _{vap}H[/tex]
Thus, we plug in the mass and the aforementioned heat of vaporization of water to obtain the following:
[tex]Q=63.7g*2259.36J/g\\\\Q=143,921J=143.9kJ[/tex]
Regards!