What rise in temperature occurs when 10g of steam at 100 degree celcius are passed into 400g of water at 10 degrees celcius taking the specific heat capacity of water to be 4.2J/gK and specific latent heat of vapourisation of steam to be 2268J/g
To find the rise in temperature when steam is passed into water, you need to calculate the heat exchanged between the steam and water.
First, let's determine the heat required to raise the temperature of the water from 10°C to the boiling point (100°C). We'll use the formula:
Q1 = m1 * c1 * ΔT1
Where:
- Q1 is the heat required
- m1 is the mass of water
- c1 is the specific heat capacity of water (4.2 J/gK)
- ΔT1 is the change in temperature
Substituting the values, we get:
Q1 = 400g * 4.2 J/gK * (100°C - 10°C)
Q1 = 168,000 J
Next, let's calculate the heat required to convert the boiling water at 100°C into steam at the same temperature. We'll use the formula:
Q2 = m2 * L
Where:
- Q2 is the heat required
- m2 is the mass of steam (which is the same as the mass of the water since the water is converted into steam)
- L is the specific latent heat of vaporization of steam (2268 J/g)
Substituting the values, we get:
Q2 = 400g * 2268 J/g
Q2 = 907,200 J
Now, let's calculate the heat required to raise the temperature of the steam from 100°C to the final temperature. We'll use the formula:
Q3 = m3 * c3 * ΔT3
Where:
- Q3 is the heat required
- m3 is the mass of steam
- c3 is the specific heat capacity of steam (which is equivalent to the specific heat capacity of water)
- ΔT3 is the change in temperature
Substituting the values, we get:
Q3 = 10g * 4.2 J/gK * (final temperature - 100°C)
Since the steam is being passed into the water until thermal equilibrium is reached, the final temperature will be the same as the initial temperature of 10°C.
Q3 = 10g * 4.2 J/gK * (10°C - 100°C)
Q3 = -3780 J
Since the heat lost by steam is equal to the heat gained by water (assuming no heat is lost to the surroundings), we can express this as:
Q1 + Q2 + Q3 = 0
168,000 J + 907,200 J - 3780 J = 0
Simplifying this equation, we get:
907,200 J + 168,000 J - 3,780 J = 0
1,071,420 J = 3,780 J
Therefore, the rise in temperature of the water is approximately 3.53°C (rounded to two decimal places).