A sample of 2.50 kg of water is held at a temperature of 100°C. How much energy must be added to completely turn the liquid water to water vapor? (The latent heat of vaporization for water is 2260 kJ/kg; the latent heat of fusion for water is 333 kJ/kg.)

Since 100C is the boiling temperature for water, for this problem we don't need to calculate the energy needed to get to the boiling point, just the heat or energy needed to vaporize the water to steam at 100C.

The formula for this is q=m(delta)

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

m is mass in grams

Delta 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.

A sample of 2.50 kg of water is held at a temperature of 100°C. How much energy must be added to completely turn the liquid water to water vapor? (The latent heat of vaporization for water is 2260 kJ/kg; the latent heat of fusion for water is 333 kJ/kg.)​

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A sample of 2.50 kg of water is held at a temperature of 100°C. How much energy must be added to completely turn the liquid water to water vapor? (The latent heat of vaporization for water is 2260 kJ/kg; the latent heat of fusion for water is 333 kJ/kg.)