Finding how much heat it takes to vaporize

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To determine the heat required to vaporize 600g of lead starting at 34°C, the initial temperature change (ΔT) is calculated as 1716°C. The formula Q = mcΔT yields a value of 133,848 J for heating the lead to its boiling point. However, it's crucial to consider the latent heat of vaporization, which must be added to the calculated heat. Additionally, the melting point of lead should be factored in, as lead may not be molten at 34°C, and the heat capacities for solid and liquid lead differ. Accurate calculations require incorporating both the heating and phase change processes.
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Homework Statement


How much heat does it take to vaporize 600g of lead that is initially at 34°C?


Homework Equations


heat capacity=1.3X102


The Attempt at a Solution


ΔT = 1750°C - 34°C
= 1716 ° C


Q= m X c X ΔT
= 0.6 X 1.3X102 X 1716
=133,848 J

Am I suppose to factor in the change of state? Is there latent vaporization?
 
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You've answered your own question. Think about the analogy of turning water to vapor.
 
Skysong12 said:
Q= m X c X ΔT
= 0.6 X 1.3X102 X 1716
=133,848 J

Am I suppose to factor in the change of state? Is there latent vaporization?

Yes you are, Q = mcΔT just gives you the heat required to raise the temperature from 34°C to 1716°C.

You just need to use Q=mlvap now and add them.
 
Skysong12 said:

Homework Statement


How much heat does it take to vaporize 600g of lead that is initially at 34°C?


Homework Equations


heat capacity=1.3X102


The Attempt at a Solution


ΔT = 1750°C - 34°C
= 1716 ° C


Q= m X c X ΔT
= 0.6 X 1.3X102 X 1716
=133,848 J

Am I suppose to factor in the change of state? Is there latent vaporization?

I not sure lead will even by molten at 34o, so you may have to melt it too. ANd is the heat capacity of solid lead and liquid lead the same?
 
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