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

Answer:

m= 5.3658 x 10³ g

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

$m=\dfrac{ 1.21268\times 10^{7}}{2.26\times 10^{6}}\ kg$

m =5.3658 kg

m= 5.3658 x 10³ g

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