Question

In exercising, a weight lifter loses 0.200 kg of water through evaporation, the heat required to evaporate the water coming from the weight lifter's body. The work done in lifting weights is 1.50 105 J. (a) Assuming that the latent heat of vaporization of perspiration is 2.42 106 J/kg, find the change in the internal energy of the weight lifter. J (b) Determine the minimum number of nutritional calories of food (1 nutritional calorie

Answer 1

a) ΔU = -6.34 × 10⁵ J

b) n = 151.46 nutritional calories

Step-by-step explanation:

Given that:

mass of the water is 0.200 kg

the work done by lifting the weight = 1.50 × 10⁵ J

Latent heat (L) = 2.42 × 10⁶ J/kg

a)

The required amount of heat Q can be calculated as:

Q = - mL

Q = (-0.200)(2.42 × 10⁶)

Q = −484000

Q = -4.84 × 10 J⁵

Now, the change in internal energy U can be calculated as:

ΔU = Q - W

ΔU = (-4.84 × 10 J⁵) - ( 1.50 × 10⁵ J)

ΔU = −634000

ΔU = -6.34 × 10⁵ J

The negative sign indicates that the internal energy is decreasing

b) Since 1 nutritional calorie = 4186 J

Then the minimum number of calories of food can be determined as:

n = ΔU / 4186 J

n = (6.34 × 10⁵ J) / 4186 J

n = 151.46 nutritional calories

Answer 2

(a) The change in internal energy of the weight lifter is -6.3 × 10⁵ J

(b) The amount of calories of food to be consumed to replace the lost energy is 151.46 calories

Step-by-step explanation:

Here we have

Mass of water, m = 0.200 kg

Latent heat of vaporization, L = 2.42 × 10⁶ J/kg

Work done in lifting the weight = 1.50 × 10⁵ J

(a) Therefore the heat required to vaporize the 0.200 kg of water is

Q = m×L = 0.200 × 2.42 × 10⁶ J/kg = -‭484,000 J ( -ve as energy is spent)

Therefore, the change in internal energy, ΔU of the weight lifter is;

ΔU = Q - W = -‭484,000 J - 1.50 × 10⁵ J = ‭-634,000 J = -6.3 × 10⁵ J

(b) The number of calories of food is given by;

1 calorie = 4186 J

Therefore, 6.3 × 10⁵ J contains

`(6.34 * 10^5)/(4186) =151.46 \, calories`

151.46 calories should be consumed to make up for the energy used.