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Ok people I have this two problems that I have an idea of how to workout but the answers I get differ from what the book says, so I need someone's help, please!!!

A 55 g ice cube at 0°C is heated until 45 g has become water at 100°C and 5 g has become steam at 100°C. How much energy was added to accomplish the transformation?

How much energy is required to heat 40g ice at -10°C to steam at 110 °C?

2007-12-03 08:56:29 · 3 answers · asked by kryptobud2003 2 in Science & Mathematics Physics

oops sorry guys its not 55g its 50g. Thanks for the heads up!!

2007-12-03 10:03:07 · update #1

3 answers

The problem with your first question is that 5g have disapeared or where not heated, well lets assume you meant 50g at the beging you have the book I don't so the latent heat of ice I don't know I will just represent it by a L

So Lf the fusion latent heat and Lg the gasification latent heat (make sure is in J/g)
the water's specific heat (I will consider is constant unless you are in advance physics its all right)
C(H2O)=1cal=4.18J/(°C*g)

so your result is

Q=50*Lf+5*Lg+100*4.18*50
sorry not to help more if you give me Lf and Lg I can do the calculus

2007-12-03 09:11:49 · answer #1 · answered by Anonymous · 0 0

You need 344 J/gm to melt ice. That takes 1 gm of ice to 1 gm of water at 0 C. So

Qmelt = 55 gm* 344J/gm = 18.92 kJ

Now you need

Qboil = m c dT of energy to take water to a boil where dT = 100 C = 100 K and c = 4.195 J/gm .

So

Qboil = 55 * 4.195*100 = 23.07 kJ

Finally you need to add 2260 J/g to convert 1 gm of water to steam. So the heat added is:

Qsteam = 5gm *2260 J/gm = 11.3 kJ

Total is Qmelt + Qboil + Qsteam = 53.3 kJ

Do teh second question the same way but then add the amount of heat needed to raise steam from 100 to 110 C. That will be given by

Q = ms Cs *dT'

where ms = mass of steam, cs = specific heat of steam (maybe different than water so look it up), and dT' = 10 C = 10 K

2007-12-03 09:06:31 · answer #2 · answered by nyphdinmd 7 · 1 0

You are missing 5 grams of water.

55 g of ice to 45 g of water and 5 grams of steam.

Either way
c = specific heat of water
clf = coefficient of latent heat of fusion
cle = coefficient of latent heat of evaporation
(I don't know these, but you book should have them)

Q = clf(m1+m2) + (m1+m2)*c*(T2-T1) + m2*cle

2007-12-03 09:04:59 · answer #3 · answered by civil_av8r 7 · 0 0

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