Question

A person who eats 3040 Cal each day consumes 1.27254 × 107 J of energy in a day. How much water at 100◦C could that much energy vaporize? The latent heat of vaporization of water is 2.26 × 106 J/

Answer 1

A person consuming 1.27254 × 10⁷ J of energy in a day could vaporize approximately 0.0563 kilograms of water at 100°C.

How can you solve how much water at 100◦C could that much energy vaporize?

First, we need to convert the person's daily calorie intake to Joules:

1 Calorie = 4.184 Joules

Therefore, 3040 Cal = 3040 * 4.184 Joules = 127254.4 Joules (rounded to 2 decimal places)

The latent heat of vaporization of water is the amount of energy required to change 1 gram of water at 100°C from a liquid to a gas (vapor). We can use this to calculate the mass of water vaporized:

Mass of water vaporized = Energy used / Latent heat of vaporization

Mass of water vaporized = 127254.4 Joules / 2.26 × 10^6 J/kg = 0.0563 kg (rounded to 3 decimal places)

Therefore, a person consuming 1.27254 × 10⁷ J of energy in a day could vaporize approximately 0.0563 kilograms of water at 100°C. This is equivalent to 56.3 grams of water.

Answer 2

5.63 gram

Step-by-step explanation:

As per the equation

Q = Lm

where Q is the amount of heat needed to convert water in to vapour

and L is the latent heat f vaporization.

m is the mass converted, hence putting all the values.

m =Q/L

= 12725400/ 2260000

= 5.63 gram