On a hot day, the temperature of a 65,000-L swimming pool increases by 1.20°C. What is the net heat transfer during this heating
1 answer:
Answer:
326149.2 KJ
Explanation:
The heat transfer toward and object that suffered an increase in temperature can be calculated using the expression:
Q = m*cv*ΔT
Where m is the mass of the object, cv is the specific heat capacity at constant volume, which basically means the amount of heat necessary for a 1kg of water to increase 1C degree in temperatur, and ΔT is the change in temperature.
A 65000 L swimming pool will have a mass of:
65000L * = 65000 kg
The specific heat capacity at constant volume of water is equal to 4.1814 KJ/KgC.
We replace the data and get:
Q = m*cv*ΔT = 65000 kg * 4.1814 KJ/KgC * 1.2°C = 326149.2 KJ
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Answer:
Compression in the spring, x = 3.7 cm
Explanation:
It is given that,
Mass of the book, m = 1.6 kg
It can be assumed the spring constant of the spring is, k = 840 N/m
As the book moves down, the change in potential energy of the book is converted to spring potential energy of compression. The mathematical expression is as follows :
x = 0.037 meters
or
x = 3.7 cm
So, the spring is compressed to a distance of 3.7 cm. Hence, this is the required solution.