Question

what will happen if 11300 kilo Joule of heat is supplied to 10 kg of water at 100°C [ given latent heat of vapourisation for water = 2260kJ/Kg]
please solve this problem
​

Answer 1

Answer:

$\red{answer}$

For a given substance, the latent heat of vaporization tells you how much energy is needed to allow for one mole of that substance to go from liquid to gas at its boiling point, i.e. undergo a phase change.

In your c no

ase, the latent heat of vaporization for water is given to you in Joules per gram, which is an alternative to the more common kilojoules per mole.

So, you need to figure out how many kilojoules per gram are required to allow a given sample of water at its boiling point to go from liquid to vapor.

As you know, the conversion factor that exists between Joules and kilojoules is

1 kJ

=103J

In your case,

2260 J/g will be equivalent to

2260Jg⋅1 kJ, 1000J=2.26 kJ/g

Now for the second part of the question. As you know,

2260=2.26⋅103

which means that2.26⋅103J=2260 J

This is the latent heat of vaporization per gram of water, which means that adding that much heat to a sample of water will vaporize one gram of water at its boiling point.

• I hope it was helpful to you :)

■■■■■■■■■■■■■■■■■■■■■■■■■■■■

$\mathfrak \green{Be Brainly}$

Answer 2

Answer:

hi oppa thank you

Explanation:

hope it will help you