Determine the quantity of heat required to vaporize 2 m3 of water at 100°C if the latent heat of vaporization of water at that temperature is 2257 kJ/kg

 Solution: To determine the quantity of heat required to vaporize 2 m³ of water at 100°C, we need to first find the mass of water and then use the latent heat of vaporization to calculate the heat required.

Given:

• Latent heat of vaporization of water at 100°C (${𝐿}_{𝑣}$) = 2257 kJ/kg
• Volume of water ($𝑉$) = 2 m³
• Density of water ($𝜌$) at 100°C = 1000 kg/m³ (approximately)

First, let's find the mass of water ($𝑚$) using the density formula:

$𝜌=\frac{𝑚}{𝑉}$

$𝑚=𝜌\times 𝑉$

$𝑚=1000\text{kg/m³}\times 2\text{m³}$

$𝑚=2000\text{kg}$

Now, we can use the formula for latent heat of vaporization to calculate the heat required ($𝑄$):

$𝑄={𝐿}_{𝑣}\times 𝑚$

$𝑄=2257\text{kJ/kg}\times 2000\text{kg}$

$𝑄=4514000\text{kJ}$

So, the quantity of heat required to vaporize 2 m³ of water at 100°C is approximately 4,514,000 kJ.