Question

On a hot day, the temperature of an 82923-L swimming pool increases by 1.6∘C. What is the net heat transfer during this heating? Ignore any complications, such as loss of water by evaporation.

Answer 1

5.5552×10⁸ J

Step-by-step explanation:

Given :

Volume of water = 82923 L

Since, 1 L = 0.001 m³

So,

Volume of water = 82.923 m³

`Density=(Mass)/(Volume)`

Density of water= 1000 kg/m³

So, mass of the water:

`Mass\ of\ water=Density * {Volume\ of\ water}`

`Mass\ of\ water=1000 kg/m^3 * {82.923 m^3}`

Mass of water = 82923 kg

Net Heat transfer during heating:

`Q=m_(water)* C_(water)* \Delta T`

Given: ΔT = 1.6 °C

Specific heat of water = 4.187 kJ/kg°C

`Q=82923* 4.187* 1.6\ kJ`

Q = 5.5552×10⁸ J