Finding how much heat it takes to vaporize

  • Thread starter Skysong12
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In summary, to vaporize 600g of lead initially at 34°C, you would need to consider both the heat required to raise the temperature from 34°C to 1716°C, as well as the latent heat of vaporization. The total heat required can be calculated using the equation Q = mcΔT + mlvap, where m is the mass of lead, c is the specific heat capacity, ΔT is the change in temperature, and lvap is the latent heat of vaporization. It may also be necessary to factor in the heat required to melt the lead if it is not already in a liquid state at 34°C.
  • #1
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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  • #2
You've answered your own question. Think about the analogy of turning water to vapor.
 
  • #3
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.
 
  • #4
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?
 
  • #5


Yes, in order to find the total amount of heat required to vaporize 600g of lead, you will need to factor in the change of state from solid to liquid to gas. This is known as the heat of vaporization. The equation for this is Q=mL, where Q is the heat required, m is the mass, and L is the specific heat of vaporization for lead. You can find the specific heat of vaporization for lead from a reliable source, such as a chemistry textbook or online database. Once you have this value, you can add it to your previous calculation to find the total amount of heat required to vaporize 600g of lead.
 

1. What is the process of vaporization?

Vaporization is the process in which a substance changes from a liquid state to a gas state. This occurs when the molecules in the liquid gain enough energy to overcome the intermolecular forces holding them together and escape into the surrounding space as a gas.

2. How is heat involved in the vaporization process?

Heat is involved in the vaporization process because it provides the energy needed to break the intermolecular bonds and convert the substance from a liquid to a gas. This energy is known as the heat of vaporization.

3. What factors determine the amount of heat needed for vaporization?

The amount of heat needed for vaporization is determined by the specific substance and its properties, such as its molecular structure, intermolecular forces, and temperature. Additionally, the pressure and surrounding environment can also affect the amount of heat required for vaporization.

4. How is the heat of vaporization measured?

The heat of vaporization is typically measured in units of energy per mole, such as joules per mole or calories per mole. This measurement can be determined through experiments that involve measuring the change in energy and temperature during the vaporization process.

5. Why is it important to know the heat of vaporization for a substance?

Knowing the heat of vaporization for a substance is important for various scientific and practical applications. It can help in predicting the behavior of substances under different conditions, such as in industrial processes or in understanding the properties of different materials. It also plays a crucial role in the design and development of technologies, such as refrigeration systems and power plants.

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