Please help me with this physics problem.?

Question

How much hear in Joules is required to boil away .5kg of water initially at 50C?

Answer 1

First the liquid must be brought to the temperature where it can start evaporating. You did not mention pressure and it is important since at atmospheric pressure different then 1 atmosphere (760mm Hg) the boiling point will vary ( see ref http://www.encyclopedia.com/doc/1E1-boilingp.html) At 1 atm the boiling point for water will be 100 degrees C.

So Qh=mk(T2-T1)
k – is specific heat of water (k=100 cal/(gm * degrees C))

Just for the record
Qh=50gm (100 cal/(gm * degrees C)) (100-50 degrees C)= 250000cal or 250kcal.

Now we have to evaporate these 50g of H2O.
Qe=mk
where k is the latent heat of evaporation for water (k=540 cal/gm)
Qe=50g (540)= 27000cal or 27 kcal

If you interested in converting cal to Joules then use 1 joule = .238845896628 cal

Qh=250000/.238845896628=104670 joules

Qe=27000/.238845896628=113044 joules

Answer 2

In this problem we are talking about water, and its boiling point is 100 degress C and it is for 1 gram. So, you start boling the water from 50 to 540. for that you need 490 degree. Because you mentioned that you need to boiled away, i am take it as empty the vessel.

use Q=mL_V therefore L_V is latent heat of vaporization, m= is 5 kg mass

You will get the answer in terms of joules......

Answer 3

To raise the water temperature requires
Q = m*c*delta T where m = .5 kg, c = 4.186 kJ / kg (degree C), and delta T = 50 degrees C (you have to heat it up to 100).

The heat of vaporization of water is 2260 kJ / kg at 100 C.