Web Analytics
Answered

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

Verified Answer :

MissPhiladelphia

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.

Answer:  \(5.4\times 10^4cal\)

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 = ?

\(Density=\frac{mass}{Volume}\)

\(1g/ml=\frac{mass}{100ml}\)

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=\(\frac{540}{1}\times 100= 5.4\times 10^4cal\)

Thus \(5.4\times 10^4cal\) must be applied to a 100 mL beaker filled with boiling water in order to vaporize the water.

Related Questions