Finding how much heat it takes to vaporize

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Homework Help Overview

The discussion revolves around calculating the heat required to vaporize 600g of lead starting from an initial temperature of 34°C. Participants are exploring the implications of phase changes and the specific heat capacities involved in the process.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are attempting to calculate the heat using the formula Q = mcΔT, but question whether they need to consider the latent heat of vaporization. There is also uncertainty about the state of lead at the initial temperature and whether it requires melting before vaporization.

Discussion Status

Some participants have provided guidance on the need to include latent heat in the calculations, while others are questioning the assumptions about the state of lead and its heat capacity. Multiple interpretations of the problem are being explored without a clear consensus.

Contextual Notes

There is uncertainty regarding the phase of lead at 34°C and whether the heat capacity values for solid and liquid lead are the same. Participants are also considering the implications of the change of state on the calculations.

Skysong12
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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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