Respuesta :
Given:
Mass of liquid = 650 g
Heat of vaporization = 723 J/g
To determine:
Heat required to vaporize 650 g of liquid
Explanation:
The heat required to vaporize 1 g of the liquid is 723 J
Therefore, the heat required to vaporize 650 g is-
= 650 g * 723 J/ 1 g = 469,950 J
Ans: Heat required is 469.95 kJ
Hello!
What is the heat required to vaporize 650 grams of a liquid with a heat of vaporization of 723 joules/gram?
We have the following data:
[tex]H_v\:(heat\:of\:vaporization) = 723\:\dfrac{J}{g}[/tex]
[tex]q\:(heat) =\:?\:(in\:Joule)[/tex]
[tex]m\:(mass) = 650\:g[/tex]
We apply the data to the formula, see:
[tex]H_v = \dfrac{q}{m}[/tex]
[tex]q = H_v * m[/tex]
[tex]q = 723\:\dfrac{J}{\diagup\!\!\!\!g} * 650\:\diagup\!\!\!\!\!g[/tex]
[tex]q = 469,950 \to \boxed{\boxed{q \approx 470,000\:J}}\end{array}}\qquad\checkmark[/tex]
Answer:
E. 470,000 J
_______________________
I Hope this helps, greetings ... Dexteright02! =)