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: [tex]5.4\times 10^4cal[/tex]
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 = ?
[tex]Density=\frac{mass}{Volume}[/tex]
[tex]1g/ml=\frac{mass}{100ml}[/tex]
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=[tex]\frac{540}{1}\times 100= 5.4\times 10^4cal[/tex]
Thus [tex]5.4\times 10^4cal[/tex] must be applied to a 100 mL beaker filled with boiling water in order to vaporize the water.
