The change in the player's internal energy is -491.6 kJ. The number of nutritional calories is -117.44 kCal
For this process to take place, some of the basketball player's perspiration must escape from the skin. This is because sweating relies on a physical phenomenon known as the heat of vaporization.
The heat of vaporization refers to the amount of heat required to convert 1g of a liquid into a vapor without causing the liquid's temperature to increase.
From the given information,
- the work done on the basketball is dW = 2.43 × 10⁵ J
The amount of heat loss is represented by dQ.
where;
∴
Using the first law of thermodynamics:b
dU = dQ - dW
dU = -mL - dW
dU = -(0.110 kg × 2.26 × 10⁶ J/kg - 2.43 × 10⁵ J)
dU = -491.6 × 10³ J
dU = -491.6 kJ
The number of nutritional calories the player has converted to work and heat can be determined by using the relation:
dU = -117.44 kcal
Learn more about first law of thermodynamics here:
brainly.com/question/3808473?referrer=searchResults
Answer:
i. 0.34
ii. 0.4
iii. 1700 w/m²
iv. 2211.36 w/m²
Explanation:
Given that
Irradiation of the plate, G = 2500 w/m²
Reflected rays, p = 500 w/m²
Emissive power, E = 1200 w/m²
See attachment for calculations
your answer is.....
D. have a large atomic radius
although they also increase going from left to right so if D is incorrect, B might be your answer. it depends on context of the lesson.