Question

What is the heat required to vaporize 650 grams of a liquid with a heat of vaporization of 723 joules/gram?

A.
0.90 J
B.
1.1 J
C.
73 J
D.
1,370 J
E.
470,000 J

Answer 1

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

Answer 2

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:  

$H_v\:(heat\:of\:vaporization) = 723\:\dfrac{J}{g}$

$q\:(heat) =\:?\:(in\:Joule)$

$m\:(mass) = 650\:g$

We apply the data to the formula, see:

$H_v = \dfrac{q}{m}$

$q = H_v * m$

$q = 723\:\dfrac{J}{\diagup\!\!\!\!g} * 650\:\diagup\!\!\!\!\!g$

$q = 469,950 \to \boxed{\boxed{q \approx 470,000\:J}}\end{array}}\qquad\checkmark$

Answer:

E. 470,000 J

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I Hope this helps, greetings ... Dexteright02! =)