Calculate the energy required to vaporize 50g of water initially at 80¡ãC. ( specific heat capacity of water- 12jgk, specific heat of vaporization- 220jk

q1 = heat needed to raise T from 80 C to 100 C.

q = mass H2O x specific heat H2O x (Tfinal-Tinitial)

q2 = heat needed to vaporize H2O at 100 to steam at 100.
q2 = mass H2O x heat vaporization.

Note: I don't agree with those numbers you have for specific heat H2O or heat vaporization for H2O.

To calculate the energy required to vaporize 50g of water initially at 80°C, we need to consider two steps: first, raising the temperature of the water from 80°C to its boiling point, and second, vaporizing the water at its boiling point.

Step 1: Calculate the energy required to raise the temperature of the water to its boiling point (100°C).

The specific heat capacity of water is given as 12 J/g°C.
The temperature change is 100°C - 80°C =20°C.
The mass of water is 50g.

Using the formula: Q = mcΔT
where Q is the energy, m is the mass, c is the specific heat capacity, and ΔT is the temperature change.

Q1 = 50g × 12 J/g°C × 20°C
= 12,000 J

So, it requires 12,000 J of energy to raise the temperature of the water from 80°C to 100°C.

Step 2: Calculate the energy required to vaporize the water at its boiling point.

The specific heat of vaporization of water is given as 220 J/g.

Using the formula: Q = mL
where Q is the energy, m is the mass, and L is the specific heat of vaporization.

Q2 = 50g × 220 J/g
= 11,000 J

So, it requires 11,000 J of energy to vaporize 50g of water at its boiling point.

The total energy required is the sum of the energy required for step 1 and step 2:

Total energy = Q1 + Q2
= 12,000 J + 11,000 J
= 23,000 J

Therefore, it requires 23,000 J of energy to vaporize 50g of water initially at 80°C.

To calculate the energy required to vaporize water, we need to consider two steps: 1) raising the temperature of water from 80°C to its boiling point, and 2) vaporizing the water at its boiling point.

Step 1: Calculating the energy to raise the temperature:
The specific heat capacity of water is given as 12 J/g°C. This means that it takes 12 Joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius. We have 50 grams of water and it needs to be raised from 80°C to its boiling point, which is 100°C.

Energy required = mass × specific heat capacity × change in temperature
= 50 g × 12 J/g°C × (100°C - 80°C)
= 50 g × 12 J/g°C × 20°C
= 12,000 J

So, it requires 12,000 Joules of energy to raise the temperature of 50 grams of water from 80°C to its boiling point of 100°C.

Step 2: Calculating the energy to vaporize the water:
The specific heat of vaporization of water is given as 220 J/g. This means it takes 220 Joules of energy to vaporize 1 gram of water.

Energy required = mass × specific heat of vaporization
= 50 g × 220 J/g
= 11,000 J

So, it requires 11,000 Joules of energy to vaporize 50 grams of water at its boiling point.

To find the total energy required, we add the energy from both steps together:
Total energy = energy to raise the temperature + energy to vaporize the water
= 12,000 J + 11,000 J
= 23,000 J

Therefore, the energy required to vaporize 50 grams of water initially at 80°C is 23,000 Joules.