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December 22, 2014

Posted by **yolanda** on Friday, April 8, 2011 at 12:59am.

- science -
**Jai**, Friday, April 8, 2011 at 2:07amto get the heat released or absorbed,

Q = mc(T2-T1)

where

m = mass of substance

c = specific heat capacity

T2 = final temperature

T1 = initial temperature

**note: if Q is (-), heat is released and if (+), heat is absorbed

now we can only apply this to substances that did not change its phase, but in the problem, we see that phase change occurs. from vapor->liquid

thus we need another data called Latent Heat of Vaporization to calculate for the heat required to change its phase:

H = m(Lv)

where

m = mass

Lv = Latent Heat of Vaporization

**note: H is (-) if heat is released and (+) if heat is absorbed

Q, total = Q1 + H + Q2

Q,total = mc(T2-T1) + mLv + mc(T3-T2)

where

T3 = 90 deg

T2 = 100 deg (boiling point of water)

T1 = 120 deg

since heat is removed, H is negative. now for every 1 kg of steam,

Q,total = 1*(c)*(100-120) - 1*Lv + 1*(c)*(90-100)

Q,total = -20*c - 10*c - Lv

Q,total = -30*c - Lv

now you look for the specific heat of water, c, and latent heat of vaporization, Lv. note that units must already be in Joules or calories.

hope this helps~ :)

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