A normal person consumes 2000 food calories a day. If a food calorie is equivalent to 4200 J of energy, use reasonable assumptions to calculate the number of kilograms of water that can be boiled away to steam starting from 20o C

The mass of water boiled away as steam is 3.24 kg

Explanation:

Given;

total food calories consumed, 2000

1 calorie = 4,200 J

total energy of food consumed in joules, Q = 2,000 x 4,200 J = 8,400,000 J

initial temperature of the water, t₁ = 20 ⁰C

Assumptions:

(1) No energy is lost

(2) all the water is boiled to steam

(3) Energy needed to raise water to 100 ⁰C (q₁)  plus  Energy needed to boil the water to steam (q₂) =  total energy of food consumed (Q)

(4) latent heat of vaporization of water, L = 2,260 kJ/kg

(5) specific heat capacity of water, c = 4,200 J/kg⁰C

let the mass of the water = m

Q = q₁ + q₂

Q = mcΔt  +  mL

Q = mc(t₂ - t₁)  +  mL

8,400,000 = 4,200(100 - 20)m  + 2,260,000m

8,400,000 = 336,000m  +  2,260,000m

8,400,000 = 2,596,000 m

m = 8,400,000/2,596,000

m = 3.24 kg

Therefore, the mass of water boiled away as steam is 3.24 kg

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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a lot of it means it can be "burned" for energy and sold at one price. if it is scarce, the price will go up