Respuesta :
Answer:
(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
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
[tex]\frac{6.34 \times 10^5}{4186} =151.46 \, calories[/tex]
151.46 calories should be consumed to make up for the energy used.
Answer:
a) ΔU = -6.34 × 10⁵ J
b) n = 151.46 nutritional calories
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
