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

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\:(J)/(g)`

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

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

We apply the data to the formula, see:

`H_v = (q)/(m)`

`q = H_v * m`

`q = 723\:(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

_______________________

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

Answer 2

The heat required to vaporize 650 grams of liquid with a heat of vaporization of 723 joules/grams is 470,000 J ( answer E)

calculation

Heat (Q) = m Hv where,

m(mass)= 650 g

HV( heat of vaporization) = 723 J/g

heat is therefore = 650 g x 723 J/g =469950 J

round off 469950 j into nearest 10,000 = 470,000 J ( answer E)