A 500 degrees Celsius heat source is being used to melt 10kg of ice, which is initially at a temperature of - 40 degrees Celsius. The heat source is transferring heat into the ice through a copper rod that is 400mm long, with a 4cm diameter. One end of the rod is in the ice.
How long will it take to melt all the ice?
After the ice has melted, how much longer (or shorter) will be the length of the copper rod, compared with the original length?
In my attempt at the first part of the question, I firstly calculated the rate of heat transfer (power) from the rod to the ice using: dQ/dt = kA dT / L, where k is the Thermal conductivity of copper, A is the cross-sectional area of the rod and L is the length of the rod.
= (400 Wm^-1K^-1)(pie(0.02)m^2)(540K-1) / 4 m) = 67.86 Watts.
I then calculated the heat required to heat the ice from - 40 degrees Celsius to its melting point (0 degrees Celsius) using Q = mc(dT) where m is the mass of the ice and c is the specific heat capacity of ice = (10kg)(2,220 J Kg^-1 K^-1)(40 K).
I then added this to the latent heat required to melt the ice: Q = mL(f), where L(f) is the specific latent heat of fusion of ice. Thus, Q = (10kg)(3.35 x 10^5J K^-1).
This gave the overall heat required to be 4,188, 000 Joules.
Finally, to get the time, I used the equation P = Q / t. Thus t = Q / P = 4,188,000 Joules / 67.86 Joules per second = 61,715 seconds. This is about 17 hours. Is this a reasonable amount of time to heat 10kg of ice?

Check your math - I think your calculation of the power transferred through the rod is off by a factor of ten. Also- it assumes that the power flow is constant, but as the ice heats from -40C to 0C the heat flow through the rod is reduced with time, which means power flow is not constant. You should end up with a differential equation that has an exponential solution. Not knowing what level of math you are expected to apply to this problem - post back and I can explain further if you think that's appropriate for your class.

Also please clarify something - in order for the heat to flow along the rod and not go out the sides the rod must be insulated, and I assume that the rod is fully inserted in the ice, right? I don't understand how the rod gets shorter?

Check your math - I think your calculation of the power transferred through the rod is off by a factor of ten. Also- it assumes that the power flow is constant, but as the ice heats from -40C to 0C the heat flow through the rod is reduced with time, which means power flow is not constant. You should end up with a differential equation that has an exponential solution. Not knowing what level of math you are expected to apply to this problem - post back and I can explain further if you think that's appropriate for your class.

Also please clarify something - in order for the heat to flow along the rod and not go out the sides the rod must be insulated, and I assume that the rod is fully inserted in the ice, right? I don't understand how the rod gets shorter?

Ok, I see now where I made the mistake when calculating the rate of heat transfer from the rod to the ice. I should have divided by 0.4 metres instead of 4 metres.
When I plug this into the power equation, this gives me a time of about 1.7 hours for the ice to melt, which is a bit more realistic.
Would you say that using this method of calculating the time for the ice to melt will always just be an approximation? That is, the heat flow will never be constant?
Can you explain further how to derive the differential equation when the heat flow is not constant?
It states in the problem that one end of the rod is in the ice (as the heat moves along the rod from the heat source to the ice), so I presume that the rod is fully insulated, although it never states this.

And what about the rod getting "shorter?" I still don't get what that's about.

The heat flow is proportional to the difference in temp between one end of the road and the other:

\frac {dQ}{dt} = \frac {kA(500-\theta)}L

where t is time and \theta is the temp of the ice. This heat flow warms the ice:

Q = mC(\theta - (-40)), or \theta = \frac Q {Mc} - 40.

Hence:


\frac {dQ} {dt} = \frac {kA(500-(\frac Q {mC} - 40))}L


To simplify things let:

a = \frac {540 kA}L \\
b = \frac {kA}{mL}


So the equation is:

\frac {dQ}{dt} = a-bQ


\int \frac {dQ}{a-bQ} = \int dt \\
- \frac {\ln(a-bQ)}b= t+c


where c is the constant of integration, which can be found fron the boundary condition that at t=0 Q=0, so :

c= \frac {- \ln(a)}b

So now we have:


t = -(c+\frac {\ln(a-bQ)} b)


You know Q from Q= mC(delta T). What you will find is that time to raise the ice temp to 0C is about 50 seconds longer than what you originally calculated.

And what about the rod getting "shorter?" I still don't get what that's about.

The heat flow is proportional to the difference in temp between one end of the road and the other:

\frac {dQ}{dt} = \frac {kA(500-\theta)}L

where t is time and \theta is the temp of the ice. This heat flow warms the ice:

Q = mC(\theta - (-40)), or \theta = \frac Q {Mc} - 40.

Hence:


\frac {dQ} {dt} = \frac {kA(500-(\frac Q {mC} - 40))}L


To simplify things let:

a = \frac {540 kA}L \\
b = \frac {kA}{mL}


So the equation is:

\frac {dQ}{dt} = a-bQ


\int \frac {dQ}{a-bQ} = \int dt \\
- \frac {\ln(a-bQ)}b= t+c


where c is the constant of integration, which can be found fron the boundary condition that at t=0 Q=0, so :

c= \frac {- \ln(a)}b

So now we have:


t = -(c+\frac {\ln(a-bQ)} b)


You know Q from Q= mC(delta T). What you will find is that time to raise the ice temp to 0C is about 50 seconds longer than what you orginally calculated.

Thanks for showing how to solve this equation - makes a lot more sense than any Physics book.
Just one thing, is the c term (for specific heat capacity) missing from b (that is, is b = kA / mcL? Just checking as it does not appear to cancel in the equation you gave for dQ / dt?
I'm not sure either what it means by the rod getting shorter either. The final part of the problem states "after the ice has melted, how much longer (or shorter) will be the length of the copper rod, compared with the original length".
I was going to used the equation for Thermal expansion where delta L = (L) (alpha L) (delta T), where alpha L is the linear expansion coefficient of copper.
Perhaps, if the rod is losing heat, some of us simpler students might be fooled into believing that it will get shorter!!
Many thanks.

Just one thing, is the c term (for specific heat capacity) missing from b (that is, is b = kA / mcL? Just checking as it does not appear to cancel in the equation you gave for dQ / dt?

Yes, you are correct - it's a mistake on my part.


I'm not sure either what it means by the rod getting shorter either. The final part of the problem states "after the ice has melted, how much longer (or shorter) will be the length of the copper rod, compared with the original length".
I was going to used the equation for Thermal expansion where delta L = (L) (alpha L) (delta T), where alpha L is the linear expansion coefficient of copper.
Perhaps, if the rod is losing heat, some of us simpler students might be fooled into believing that it will get shorter!!!
Many thanks.

Ahh.. I hadn't thought about the thermal expansion of the rod as it heats. In the beginning the temp varies from 500C at one end to -40C at the other. You can assume a linear temp profile and then integrate using the coefficient of expansion to find the total change in length compared to its final length when the temp profile is 500C at the hot end and 0C at the cold end.

Yes, you are correct - it's a mistake on my part.



Ahh.. I hadn't thought about the thermal expansion of the rod as it heats. In the beginning the temp varies from 500C at one end to -40C at the other. You can assume a linear temp profile and then integrate using the coefficient of expansion to find the total change in length compared to its final length when the temp profile is 500C at the hot end and 0C at the cold end.

Sorry, I was just wondering if you could clarify something.
Should Q be equal to mc (theta - (-40)) + mLf, where m is the mass of the ice and Lf is the specific latent heat of fusion of ice - I highlighted the extra bit) - because they ask in the question how long it takes to melt the ice (not just to heat it up to 0 degrees)
If this is true, how would it alter the differential equation?

Sorry, I was just wondering if you could clarify something.
Should Q be equal to mc (theta - (-40)) + mLf,

That would be the total heat required to complete the process of melting the ice, yes. My previous two posts were about the first phase of the process, as the ice warms to 0C. The second phase - melting the ice - I think you had already done correctly, since during that phase the temp of the rod and the heat flow through it remains constant and there is need for a differential equation for that part.