How many kcal of heat are required to vaporize 25.0kg of water at 100°C?
To determine the number of kilocalories (kcal) of heat required to vaporize water, we need to use the equation:
Q = m * ΔH
Where:
Q = heat energy in kcal
m = mass of the substance in kg
ΔH = heat of vaporization in kcal/kg
The heat of vaporization of water at 100°C is approximately 540 kcal/kg.
Now, let's calculate the heat energy required to vaporize 25.0 kg of water at 100°C:
Q = 25.0 kg * 540 kcal/kg
Q = 13,500 kcal
Therefore, it takes 13,500 kcal of heat to vaporize 25.0 kg of water at 100°C.
To calculate the amount of heat required to vaporize water, we need to use the heat of vaporization (also known as latent heat) of water. The heat of vaporization of water is approximately 540 kcal/kg.
Given:
Mass of water (m) = 25.0 kg
Heat of vaporization (Hv) = 540 kcal/kg
The formula to calculate the heat required (Q) is:
Q = m * Hv
Substituting the values:
Q = 25.0 kg * 540 kcal/kg
Q = 13,500 kcal
Therefore, it requires 13,500 kcal of heat to vaporize 25.0 kg of water at 100°C.
Q=λ•m
1 J =0.2388 Cal
Heat of vaporization for water
λ = 2257•10^3•0.2388 =5.39•10^5 Cal/kg,
Q=λ•m =5.39•10^5•25 =1.35•10^7 Cal =
= 1.35•10^4 kCal