Question

On a hot day, the temperature of an 80,000-L swimming pool increases by 1.50°C. What is the net heat transfer during this heating? Ignore any complications, such as loss of water by evaporation.

(a) 1.2 x 10⁹ J
(b) 1.8 x 10⁹ J
(c) 2.4 x 10⁹ J
(d) 3.0 x 10⁹ J

Answer 1

The net heat transfer during the heating of an 80,000-L swimming pool can be calculated using the formula Q = mc ΔT, where Q is the heat transfer, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature. Using the given values, the net heat transfer is 501,600,000 J.

Step-by-step explanation:

The net heat transfer during the heating of an 80,000-L swimming pool can be calculated using the equation:

Q = mcΔT

Where Q is the heat transfer, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.

In this case, the mass of the water can be calculated using the density of water:

m = ρV

Where ρ is the density of water and V is the volume of the pool.

Using the specific heat capacity of water (4.18 J/g°C) and the density of water (1 g/mL), we can calculate the heat transfer as:

Q = (ρV)(c)(ΔT)

Substituting the values given in the question:

Q = (1 g/mL * 80,000 L)(4.18 J/g°C)(1.50°C)

Converting mL to g and L to mL:

0 g)(4.18 J/g°C)(1.50°C)

Simplifying this expression gives the net heat transfer during the heating of the swimming pool as:

Q = 501,600,000 J