Question

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

Answer 1

Answer: 469,950 J

Explanation:

Vaporization heat is the amount of heat energy needed to vaporize an amount of substance at a constant temperature, this is to change the liquid into gas phase. The temperature remains constant during the change of phase because the energy is being used to beat the intermolecular forces that exist in the liquid phase and so the particles (molecules) can separate to change the state from liquid to gas.

The specific vaporization heat is given in units of joules per gram, meaning that to compute the total amount of heat you just need to multiply the specific vaporization heat times the amount of substance:

Heat energy = mass in grams × specific vaporization heat

Heat energy = 650 g × 723 J/g = 469,950 J ← answer,

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:

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

$q = H_v * m$

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

$q = 469950 \to \boxed{\boxed{q \approx 470000\:J}}\end{array}}\qquad\checkmark$

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