Question

A person who eats 2897 Cal each day consumes 1.21268 × 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/kg. Answer in units of g

Answer 1

To find the amount of water that could be vaporized, we use the equation Qv = mLv, where Qv is the energy needed to evaporate mass m and Lv is the latent heat of vaporization. Substituting the given values, we find that the person could vaporize approximately 5.372566 kg of water at 100°C.

Step-by-step explanation:

To calculate the amount of water that could be vaporized by a person consuming 2897 Cal (1.21268 × 10^7 J) of energy in a day, we need to find the number of kilograms of water that corresponds to that energy. The latent heat of vaporization of water is 2.26 × 10^6 J/kg. We can use the equation Qv = mLv, where Qv is the energy needed to evaporate mass m and Lv is the latent heat of vaporization. Rearranging the equation, we find m = Qv / Lv. Substituting the given values, we have m = (1.21268 × 10^7 J) / (2.26 × 10^6 J/kg). Solving for m, we get m = 5.372566 kg.

Therefore, the person could vaporize approximately 5.372566 kg of water at 100°C.

Answer 2

m= 5.3658 x 10³ g

Step-by-step explanation:

Given that

Energy eat each day = 2897 cal

Energy consume each day ,Q = 1.21268 x 10⁷ J

Latent heat ,LH = 2.26 x 10 ⁶ kg/J

Lets take , the mass of the water vaporize = m kg

From energy balance

Q = m x LH

1.21268 x 10⁷ = 2.26 x 10 ⁶ x m

`m=( 1.21268* 10^(7))/(2.26* 10^(6))\ kg`

m =5.3658 kg

m= 5.3658 x 10³ g

Therefore the mass of the water is 5.3658 x 10³ g.