Question

1. on a hot day, the temperature of an 80,000-l swimming pool increases by 1.50 oc. what is the net heat transfer during this heating? ignore any complications, such as loss of water by evaporation.

Answer 1

To calculate the net heat transfer of an 80,000-L swimming pool that has increased by 1.50°C, use the heat transfer equation with the specific heat capacity of water. The resulting heat transfer is 502.8 megajoules.

Step-by-step explanation:

Calculating Net Heat Transfer for a Swimming Pool

To determine the net heat transfer for the 80,000-L swimming pool with a temperature increase of 1.50°C, one must use the specific heat capacity formula.

The formula is Q = mcΔT, where 'Q' is the heat transfer, 'm' is the mass of the water, 'c' is the specific heat capacity of water (approximately 4.18 J/g°C), and 'ΔT' is the change in temperature.

Assuming the density of water is 1 kg/L, the mass 'm' of the 80,000 L of water is 80,000 kg.

Given that the temperature change 'ΔT' is 1.50°C, the heat transfer 'Q' can be calculated as follows:

Q = (80,000 kg) × (4.18 J/g°C × 1000 g/kg) × (1.50°C)

= (80,000 × 4.18 × 1.50) × 1000 J

Q = 502,800,000 J

Therefore, the net heat transfer during the heating of the pool by 1.50°C is 502.8 MJ (megajoules).