How much thermal energy must be added to 0.96 kg of water at 100 ∘C to make steam at 100 ∘C?
from google search
The heat of vaporization of water is about 2,260 kJ/kg
what is it in a decimal
Well, to make steam at 100 ∘C, you must heat up the water. And we all know water is a little bit of a diva. It likes to stay at 100 ∘C and only then, it's ready to turn into steam. So, you'll need to add enough thermal energy to keep it at that temperature. Just make sure to use a good amount of humor to keep the water entertained. It'll turn into steam before you know it! Just be careful not to make it too hot, or you might end up with a fire-breathing dragon instead of steam.
To calculate the thermal energy required to convert water at 100°C to steam at 100°C, we need to use the equation:
Q = m * L
Where:
Q is the thermal energy
m is the mass of the substance (in this case, water)
L is the latent heat of vaporization
The latent heat of vaporization for water is 2.26 x 10^6 J/kg.
Given:
m = 0.96 kg
Using the equation, we can calculate the thermal energy needed:
Q = 0.96 kg * 2.26 x 10^6 J/kg
Q ≈ 2.17 x 10^6 J
Therefore, approximately 2.17 x 10^6 joules of thermal energy must be added to convert 0.96 kg of water at 100°C to steam at 100°C.
To find the amount of thermal energy required to convert water at 100°C to steam at 100°C, we need to consider two steps:
1. Raising the temperature of the water from 100°C to its boiling point (100°C).
2. Vaporizing the water at its boiling point (100°C) to steam at the same temperature.
Step 1: Raising the temperature of the water
The specific heat capacity of water is 4.18 J/g⋅°C. Therefore, we need to calculate the thermal energy required to raise the temperature of the water from 100°C to its boiling point.
Since we have 0.96 kg of water, we can convert it to grams by multiplying by 1000.
0.96 kg * 1000 g/kg = 960 g
The temperature change is:
ΔT = 100°C - 100°C = 0°C
The thermal energy required is given by:
Q1 = mass * specific heat capacity * ΔT
Q1 = 960 g * 4.18 J/g⋅°C * 0°C
Q1 = 0 J
Step 2: Vaporizing the water to steam
The heat of vaporization of water is 2260 J/g, which is the amount of thermal energy required to convert 1 gram of water at its boiling point to steam at the same temperature.
To calculate the thermal energy required to vaporize the water, we use:
Q2 = mass * heat of vaporization
Q2 = 960 g * 2260 J/g
Q2 = 2177600 J
The total thermal energy required is the sum of Q1 and Q2:
Total thermal energy = Q1 + Q2
Total thermal energy = 0 J + 2177600 J
Total thermal energy = 2177600 J
Therefore, the amount of thermal energy required to convert 0.96 kg of water at 100°C to steam at 100°C is 2177600 J.