Latent Heat

Question

I've worked on this one for hrs and finally quit.

 

How much heat must be added to .45kg of aluminum to change it from a solid at 130C to a liquid at 660C? The latent heat of fusion for aluminum is 4.0E5 J/kg.

The book answer is 3.9E5 but I believe it should be 2.14E5

 

thank you

Answer 1

John,

The amount of heat energy you need to raise the temperature of the aluminum to its melting temperature is

Q₁ = cmΔT, where c = specific heat constant for AL = 0.91

Q₁ = (0.91 KJ/Kg^oC) (0.45 Kg) (660 - 130 ^oC) = 217 KJ

 

And the amount of additional heat energy needed to melt it is 

Q₂ = mL, L = latent heat coefficient for AL

Q₂ = (0.45 Kg) ( 400K/Kg) = 180 KJ

 

So total required heat energy = Q₁ + Q₂ = 217 + 180 = 397 KJ = 3.97 E5 Joules

 

That's slightly off from your 3.9 E5 figure.  Did you forget to type the 7 or maybe your constants are slightly different.

Answer 2

What is the melting point for aluminum? To solve this you use the specific heat from the temperature to the melting point, then the latent heat then again the specific heat. For example to get 1 gram of water at 20 deg C to 1 gram of steam at 100 C it takes:

 

(100-20)*specific heat+latent heat