Posted by Matt W on Wednesday, January 5, 2011 at 10:52am.
I did this problem last night with Damon's help and I think it's correct at the end of the iscould you please check it?
Calculate amount of energy needed to completely vaporize a 50g piece of 20 degree C silver
melting point 1064 C
L(f) = 64KJ/Kg
Boiling point 2856C
L(v) = 1645 KJ/Kg
c= 0.13KJ/Kg degrees C
I think I take the (2856  20) * 50g *0.13 = 184.34kJ
Then I take 1064  20 * 50 g * 0.13 = 67.87
184.34/67.87 =17.40 kJ
Thank you
PhysicsPlease recheckI've attempted it  Matt W, Tuesday, January 4, 2011 at 6:51pm
So if I did the corrections, then it would equal 18.434
and the second calculation would equal 6.786
.18434/.06786=.270 KJ
Correct, Better or still wrong
PhysicsPlease checkI've attempted it  Damon, Tuesday, January 4, 2011 at 6:53pm
first heat it up from 20 to 1064
(1064 20)(.13)(.05) = a
now melt it
64 (.05) = b
now heat it from 1064 to 2856
(28561064)(.13)(.05) = c
now boil it
1645(.05) = d
total heat = a+b+c+d
PhysicsIs this correct?  Matt W, Tuesday, January 4, 2011 at 7:33pm
a= 1044 * .13 * .05 = 6.786
b= 64 * .05 = 3.2
c= 1792 * .13*.05 = 11.648
d= 1645 * .05 =82. 25
6.786 + 3.2 + 11.648 + 82.25 = 103.884

PhysicsPlease check calc  drwls, Wednesday, January 5, 2011 at 12:33pm
It looks correct but I was surprised and skeptical about the specific heats of solid and liquid being the same (0.05 cal/g deg C).
According to one website I found,
".. silver's specific heat capacity as a solid is 0.235 kilojoules per kilogram deg C. As a liquid, it's 0.278 kilojoules per kg deg C." Those would correspond to 0.056 and 0.066 cal/g deg C
The answer you'd get using those specific heat values would be only slightly different
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