Time for melting and vaporising a wax candle

In summary, the problem involves calculating the time it takes for a 50g wax candle at 25 degrees Celsius to melt and vaporize under constant heat admission. Using the given data and equations for heat of transformation, specific heat capacity and power, the steps for melting and vaporization are determined and used to calculate the total time. The final equation used is Δt=Q/P, where Q is the total heat and P is the heat admission rate.
  • #1
kotchenski
16
0

Homework Statement


Assume that a wax candle with mass m at temperature T1 undergoes constant heat admission dQ/dt=Kp

Assume:
m=50g
T1=25 celsius
Kp=10W

Calculate how long it will take for the wax to melt and vaporise during constant heat admission
without any loss.

Data:
ρ=791 kg/m3
M=310 g/mol
Tm=317.15 K (melting temperature)
Lf=252 kJ/kg
Tb=641.8 K (boiling temperature)
Lv=105 kJ/kg
Cp(s)=598.1 J/mol K
Cp(l)=739 J/mol K
Cp(g)=1193 J/mol K

Homework Equations


P=Q/Δt

Q=mL
Q=mcΔT

The Attempt at a Solution


Well the posibilities overwhelmes me.

I know that it all boils down to using Δt=Q/P the question is though which Q are they asking for, I have only made an attempt at the boiling question though.

For boiling
If I apply what I know from heat of transformation:
Q=mL
Q=0.05kg*252 kJ/kg -> Q=12600J
Thereby:
Δt=12600J/10W -> Δt=1260s
In other words it would take 21mins.

However if that's the case then it would be even faster to boil it away when it's in liquid.

Am I doing something wrong or is there something blatantly obvious that I don't seem to get?
 
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  • #2
kotchenski said:
Lf=252 kJ/kg

For boiling
If I apply what I know from heat of transformation:
Q=mL
Q=0.05kg*252 kJ/kg -> Q=12600J
Is this consistant?

Before plugging any numbers into equations, can you describe to me what will happen to the candle, from beginning to end?
 
  • #3
DrClaude said:
Is this consistant?

Before plugging any numbers into equations, can you describe to me what will happen to the candle, from beginning to end?

I can see I said I would calculate the boiling, when I mean't I would calculate the melting.

The candle is in solid phase and it's placed in a isolated room which has the temperature of 25 celsius. I will now light the candle and have to calculate the time it will take the 50g of solid wax to become liquid.

I tried to calculate the time it would take for the candle to melt, and my guess for that was to use the equation Q=mL.

I here apply the constant Lf since the candle is going from solid to liquid state. The heat I calculate from that is 12.6 kJ or 12600J.

From that I then use Δt=Q/P

What I'm unsure about is whether or not I've done it correctly given the options that I have.
 
  • #4
kotchenski said:
I can see I said I would calculate the boiling, when I mean't I would calculate the melting.
Ok.

kotchenski said:
The candle is in solid phase and it's placed in a isolated room which has the temperature of 25 celsius. I will now light the candle
You are not lighting the candle. You are supplying heat at a uniform rate. It is not the same thing.

kotchenski said:
and have to calculate the time it will take the 50g of solid wax to become liquid.
But what happens before it becomes liquid?

kotchenski said:
I tried to calculate the time it would take for the candle to melt, and my guess for that was to use the equation Q=mL.
That part is correct. But if you want to get through the problem, you first have to figure out everything that happens from beginning to end.
 
  • #5
DrClaude said:
Ok.


You are not lighting the candle. You are supplying heat at a uniform rate. It is not the same thing.


But what happens before it becomes liquid?


That part is correct. But if you want to get through the problem, you first have to figure out everything that happens from beginning to end.

I assume there are two stages then, the first stage is when the wax is heating which is expressed by:
[itex]Q= \frac{m C_p (s) (T_m-T_1)}{M}[/itex]

The second stage would be when the wax reaches the melting temperature and begins to melt and so I add that to the equation so that Q will be:
[itex]Q= \frac{m C_p (s) (T_m-T_1)}{M} +mL_f[/itex]

From this I can then apply my final formula:
[itex]\Delta t= \frac{Q}{P}[/itex]

Would this be correct?
 
  • #6
kotchenski said:
I assume there are two stages then, the first stage is when the wax is heating which is expressed by:
[itex]Q= \frac{m C_p (s) (T_m-T_1)}{M}[/itex]

The second stage would be when the wax reaches the melting temperature and begins to melt and so I add that to the equation so that Q will be:
[itex]Q= \frac{m C_p (s) (T_m-T_1)}{M} +mL_f[/itex]
This is correct but incomplete. The problem asks for the time for complete vaporization.
 
  • #7
DrClaude said:
This is correct but incomplete. The problem asks for the time for complete vaporization.

Not entirely. I'm sorry if the translation is terrible but it's in danish. They ask for the time it takes for it to melt and then to vaporize. For vaporization I assume it would be in 4 stages:

1. Heating to reach the melting temperature.
2. Melting the wax.
3. Heating the liquid wax to the boiling temperature.
4. The wax is vaporizing.

This would give the heat calculation Q:

[itex]Q= \frac{m C_p (l) (T_b - T_m)}{M} + mL_v + \frac{m C_p (s) (T_m - T_1)}{M} + mL_f[/itex]

And then it is just [itex] \Delta t = \frac{Q}{P}[/itex]

Now I should be home free unless there is more to it.
 
  • #8
kotchenski said:
Not entirely. I'm sorry if the translation is terrible but it's in danish. They ask for the time it takes for it to melt and then to vaporize. For vaporization I assume it would be in 4 stages:

1. Heating to reach the melting temperature.
2. Melting the wax.
3. Heating the liquid wax to the boiling temperature.
4. The wax is vaporizing.

This would give the heat calculation Q:

[itex]Q= \frac{m C_p (l) (T_b - T_m)}{M} + mL_v + \frac{m C_p (s) (T_m - T_1)}{M} + mL_f[/itex]

And then it is just [itex] \Delta t = \frac{Q}{P}[/itex]

Now I should be home free unless there is more to it.
Perfekt!
 
  • #9
DrClaude said:
Perfekt!

You purposely searched up that word didn't ya? :D, regardless thanks for your help!
 
  • #10
kotchenski said:
You purposely searched up that word didn't ya? :D
Nej, jag kan svenska.

kotchenski said:
regardless thanks for your help!
You're welcome!
 

Related to Time for melting and vaporising a wax candle

1. How does temperature affect the melting and vaporizing time of a wax candle?

The higher the temperature, the faster the wax candle will melt and vaporize. This is because heat causes the molecules in the wax to move faster, resulting in a higher rate of evaporation.

2. Does the type of wax used in a candle affect the time it takes to melt and vaporize?

Yes, different types of wax have different melting and vaporizing points. For example, paraffin wax has a lower melting point compared to beeswax, so it will melt and vaporize at a faster rate.

3. How does the size and shape of a candle impact the melting and vaporizing time?

The size and shape of a candle can affect its surface area and therefore, the rate at which it melts and vaporizes. A larger candle with a larger surface area will melt and vaporize faster compared to a smaller candle with a smaller surface area.

4. Is there a difference in the melting and vaporizing time between a lit and unlit candle?

Yes, a lit candle will melt and vaporize faster compared to an unlit one. This is because the heat from the flame accelerates the melting and vaporizing process.

5. How can factors such as humidity and air flow affect the melting and vaporizing time of a candle?

Humidity can slow down the melting and vaporizing process of a candle as the moisture in the air can prevent the wax from evaporating. On the other hand, increased air flow can speed up the process as it helps to carry away the vaporized wax particles.

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