Question

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? Ignore any complications, such as loss of water by evaporation. (Enter the magnitude.)

Answer 1

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 *$\frac{1m^3}{1000L} * \frac{1000kg}{1m^3}$ = 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