How much energy is necessary to heat 5 grams of water from 27 degrees to 50 degrees C?

Question

1. Calculate the energy necessary to heat 5 grams of water from 27 degrees to 50 degrees C.

This is also one I don't understand...

2. Calculate the energy necessary to boil a cup of water at 27 degrees C to steam at 100 degrees C. ( 1 cup is 236.5 grams of water)

Answer 1

The specific heat of water is 1 cal/g°K by definition and the latent heat of vaporization of water is 540 cal/g at 100 C.

1. 5 x (50-27) = 115 calories

2. First heat the water to 100°C, then boil it.

236.5 x (100-27) + 236.5 x 540 = 17264.5 + 127710 = 144674.5 cal.

Answer 2

1 calorie is by definition the amount of energy required to increase the temperature of 1 gram of water by 1 degree C. You have 5 grams, and you need to increase its temperature by 23 degree, so 5*23=115 calories (a calorie is 4.184 joules, so if you prefer the answer in that unit, then it is 481.16 joules)

Boiling water is however a two step process. First you have to heat the water to 100 C, and then you have to add more energy to turn the water at 100 C into steam at 100 C. This heat of vaporisation is equal to 2270 joules/g.

So, 236.5 g with a variation of 73 degree: 17264.5 calorie or 72234.7 Joules.

236.5 g with a vaporisation energy of 2270 g: 536855 joules.

Total: 609089.7 joules.