SOLUTION: how much heat is required to convert 50g of ice at -10degree celcius to steam at 200degree celcius?

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Question 826833: how much heat is required to convert 50g of ice at -10degree celcius to steam at 200degree celcius?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
How much heat is required to convert 50g of ice at -10%5Eo Celsius to steam at 200%5Eo Celsius?

This is a problem that a science teacher (chemistry or physics) would give you.
You need to assume that the process occurs at standard atmospheric pressure (1 atm). (The pressure at start and end is really what matters, and we should assume that is is 1 atmosphere).
You need to calculate 5 different amounts of heat for 5 different stages of the process:
1) the heating of ice from r to 0%5EoC ,
2) the melting at 0%5EoC ,
3) the heating of the resulting liquid water from 0%5EoC to 100%5EoC ,
4) the vaporization of the liquid water at 100%5EoC , and
5) the heating of the resulting water vapor from 100%5EoC to 200%5EoC .
You need data from your textbook, or from a handbook of physics (or chemistry, or engineering).
I tried to find data, and calculate accurately, but you may have more accurate data, or be less prone to calculation mistakes, or your data may be in different units.

Water, as a liquid between 0%5EoC and 100%5EoC ,
has a "specific heat" of about 1calorie%2Fg%2FdegreeC .
As ice, a bit below 0%5EoC , or
as vapor at 100%5Eo to 200%5EoC ,
it has a "specific heat" of about 0.5calorie%2Fg%2FdegreeC .
That is the heat it takes to increase the temperature of 1g water by 1%5EoC (according to my handbook).
(Of course, the number is a function of temperature, but it does no vary by too much).
The heats, or "latent heats", of melting (for ice) and vaporization (for liquid water) are about 1440 and 9700calorie%2Fmol respectively (according to my handbook).

So the amounts of heat required are
%2850g%29%2A0.5cal%2Fg%280-%28-10%29%29%5EoC=250cal to heat the ice to 0%5EoC ,
%2850g%2F%2218+g+%2F+mol%22%29%2A1440calorie%2Fmol=4000cal= about 4cal to melt the ice at 0%5EoC ,
%2850g%29%2A1cal%2Fg%28100-0%29%5EoC=5000cal to heat the liquid water from 0%5EoC to 100%5EoC ,
%2850g%2F%2218+g+%2F+mol%22%29%2A9700calorie%2Fmol=27000cal to vaporize the water at 100%5EoC , and
%2850g%29%2A0.5cal%2Fg%28200-100%29%5EoC=2500cal to heat the water vapor from 100%5EoC to 200%5EoC .
According to my calculation, the total energy in calories is
250%2B4000%2B5000%2B27000%2B2500=38700 .
NOTE:
Your data may slightly different, or it may be in different units (Joules rather than calories, or Joule/g rather than calorie/mol, for example).