How much heat (calories) must be applied to a 100 mL beaker filled with boiling water in order to vaporize the water?
** A) 540 cal **
B) 540 * 10^4 cal
C) 800 cal
D) 8000 cal
Is A correct? or cud it be B.?
You will need 540 calories per gram
water is 1 mL or 1 cm^3 per gram
so I get
540 * 100 csl
which is
5.40 * 10^4 cal
I hope B is a typo and should be 5.40*10^4
thnx again!
typo meaning?... the "cal" at the end?
you had typed 540*10^4
it should be
5.40 * 10^4
oops! ur right... typo there.
5.40*10^4
To find the amount of heat required to vaporize the water, you can use the formula:
Q = m * ΔHv
Where:
Q is the amount of heat (calories) required
m is the mass of the substance (in this case, water)
ΔHv is the heat of vaporization
To calculate the mass of water in the beaker, you can use the density of water:
Density = mass / volume
The density of water is approximately 1 g/mL.
Therefore, the mass of 100 mL of water is 100 g.
The heat of vaporization for water is approximately 540 calories/g.
Plugging the values into the formula:
Q = 100 g * 540 cal/g
Q = 54,000 cal
So, the correct answer is A) 540 cal.