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 is-could 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

Physics-Please recheck-I'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

Physics-Please check-I'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

(2856-1064)(.13)(.05) = c

now boil it

1645(.05) = d

total heat = a+b+c+d

Physics-Is 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

- Physics-Please 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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