a sample of 2.50kg of water is held at a temperature of 100 celsius. how much energy must be added to completely turn the liquid water to water vapor

Answer 1

Look up the latent heat of vaporization (l_v) of water, preferably in units of kJ/kg. (There should be a table of latent heats in your physics book.) Multiply the given mass of water (in kg) by the latent heat of vaporization (in kJ/kg). That gives you the required energy transfer (ΔE) into the liquid water.

m = 2.50 kg

l_v = ? kJ/kg (look it up)

ΔE = ml_v = ? kJ

Answer 2

Since 100C is the boiling temperature for water, for this problem we dont need to calculate the energy needed to get to boiling point, just the heat or energy needed to vaporize the water to steam at 100C.

The formula for this is q=m(deltaH)

q is Joules of heat needed to vaporize the water to steam at 100C

m is mass in grams

deltaH is in Joules per gram and can be looked up for water at this temperature. Here, it is approximately 2260J/g. This online lecture should help ease understanding: https://cabrillo.instructure.com/courses/10267/modules/items/256219

Therefore...

q=2.5g (2260J/g)= 5650J = 5.65kJ

I do not do Physics tutoring but am happy to answer questions here.