Question

How much heat (in calories) must be applied to a 100 mL beaker filled with boiling water in order to vaporize the water?

540
5.4 x 104
800
8,000

Answer 1

Answer:
`5.4* 10^4cal`

Step-by-step explanation:

Latent heat of vaporization is the amount of heat required to convert liquid to gas at atmospheric pressure.

Latent heat of vaporization = 540 cal/g

Volume of water = 100 ml

Density of water = 1 g/ml

Mass of water = ?

`Density=(mass)/(Volume)`

`1g/ml=(mass)/(100ml)`

Mass of water = 100 g

Amount of heat required to vaporize 1 g of boiling water = 540 cal

Amount of heat required to vaporize 100 g of boiling water=
`(540)/(1)* 100= 5.4* 10^4cal`

Thus
`5.4* 10^4cal` must be applied to a 100 mL beaker filled with boiling water in order to vaporize the water.

Answer 2

The latent heat of vaporization of water is 540 cal/g/°C. The amount of water is 100 mL. Since the density of water is 1 g/mL, this is equivalent to 100 g. Thus,

540 cal/g/°C * 100 g = 54,000 cal

The answer is 5.4 x 10^4 cal.