Time for melting and vaporising a wax candle

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SUMMARY

The discussion focuses on calculating the time required for a wax candle to melt and vaporize under constant heat admission of 10W. The candle has a mass of 50g and starts at a temperature of 25°C. The calculations involve determining the heat required for both melting (using latent heat of fusion, Lf = 252 kJ/kg) and vaporization (using latent heat of vaporization, Lv = 105 kJ/kg), along with the specific heat capacities for solid, liquid, and gas phases. The final formula used to calculate the time is Δt = Q/P, where Q accounts for the total heat required for both processes.

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  • Understanding of thermodynamics, specifically heat transfer and phase changes.
  • Familiarity with specific heat capacities (Cp) for solid, liquid, and gas phases.
  • Knowledge of latent heat concepts, including latent heat of fusion and vaporization.
  • Ability to apply the formula Δt = Q/P for calculating time based on heat and power.
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  • Study the principles of phase transitions and how they affect energy calculations in thermodynamics.
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kotchenski
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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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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 consistent?

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

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.
 
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.
 
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:
Q= \frac{m C_p (s) (T_m-T_1)}{M}

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:
Q= \frac{m C_p (s) (T_m-T_1)}{M} +mL_f

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

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

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:
Q= \frac{m C_p (s) (T_m-T_1)}{M} +mL_f
This is correct but incomplete. The problem asks for the time for complete vaporization.
 
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:

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

And then it is just \Delta t = \frac{Q}{P}

Now I should be home free unless there is more to it.
 
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:

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

And then it is just \Delta t = \frac{Q}{P}

Now I should be home free unless there is more to it.
Perfekt!
 
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!
 

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