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

Answer:

5.5552×10⁸ J

Explanation:

Given :

Volume of water = 82923 L

Since, 1 L = 0.001 m³

So,

Volume of water = 82.923 m³

$Density=\frac{Mass}{Volume}$

Density of water= 1000 kg/m³

So, mass of the water:

$Mass\ of\ water=Density \times {Volume\ of\ water}$

$Mass\ of\ water=1000 kg/m^3 \times {82.923 m^3}$

Mass of water  = 82923 kg

Net Heat transfer during heating:

$Q=m_{water}\times C_{water}\times \Delta T$

Given: ΔT = 1.6 °C

Specific heat of water = 4.187 kJ/kg°C

$Q=82923\times 4.187\times 1.6\ kJ$

Q = 5.5552×10⁸ J