Calculating Mass of Melting Ice in a Copper Rod | Introductory Physics Homework

In summary, this problem asks you to find out how much ice melts per second when one end of a copper rod is in contact with boiling water and the other in a mixture of water and ice. You assume no heat is lost through the side of the rod, and find that the ice melts at a rate of 3.1045*10^-5 kg/s.
  • #1
2099
4
0
Hello,
This is a homework problem for my Introductory Physics(no calculus) class but it's relatively simple so I opted to put it under high school. First I'll state the problem then my thoughts.

A copper rod (k = 390) has a length of 1.5 m and a cross-sectional area of 4.00*10^-4 m^2. One end of the rod is in contact with boiling water and the other with a mixture of water and ice. What is the mass of ice per second that melts? Assume that no heat is lost through the side surface of the rod.

This seems simple enough. Obviously
Q=(390*(4*10^-4)*DT*t)/1.5
DT is where I am coming across a problem. Am I supposed to assume that it is 100-0? Doing so gives
Q=10.4*t J*s
My idea is that in order for the ice to melt it has to have at least m*L (L=33.5*10^4) so I set 10.4*t=m*L
and it follows m/t=10.4/(33.5*10^-4) which is approximately 3.1045*10^-5 kg/s

I would greatly appreciate it if any could help me with this. I can't seem to find much information on this type of problem and I've looked to other books. They seem to have this problem, but no solution.
Anyway, I'm not looking for an answer just trying to see if my logic is correct. Thank you in advance
 
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  • #2
2099 said:
A copper rod (k = 390) has a length of 1.5 m and a cross-sectional area of 4.00*10^-4 m^2. One end of the rod is in contact with boiling water and the other with a mixture of water and ice. What is the mass of ice per second that melts? Assume that no heat is lost through the side surface of the rod.

This seems simple enough. Obviously
Q=(390*(4*10^-4)*DT*t)/1.5
DT is where I am coming across a problem. Am I supposed to assume that it is 100-0? Doing so gives
Q=10.4*t J*s
My idea is that in order for the ice to melt it has to have at least m*L (L=33.5*10^4) so I set 10.4*t=m*L
and it follows m/t=10.4/(33.5*10^-4) which is approximately 3.1045*10^-5 kg/s
Since the question asks for the rate of ice mass melting/second, you are interested in finding Q/t. This is proportional to the thermal conductivity, area and temperature difference and inversely proportional to the length of the conducting body (rod). You have to use the entire temperature difference, which is 100 degrees. That gives you Q/t of 10.4 J/sec. as you have found, which looks correct.

As you seem to realize, you have to figure out how much heat flow is required to melt one gram or kg. of ice per second. The latent heat of water is 334 kJ./kg. or 334 J/g.

The rest is just plugging in the numbers.

AM
 
  • #3
Thank you for your reply. I understand what you are saying I just can't tell if I've set it up correctly or not. There is 10.4 J/s or W going to the mixture, I know it takes 33.5*10^4 J to melt 1 kg of ice, but it is the time factor that is confusing me. Is this Q/t=(mL)/t okay, because if I don't put the time on the other side I get something like kg = kg/s
Anyway, I still come up with 3.1045*10^-5 kg/s approximately.
 
  • #4
2099 said:
Thank you for your reply. I understand what you are saying I just can't tell if I've set it up correctly or not. There is 10.4 J/s or W going to the mixture, I know it takes 33.5*10^4 J to melt 1 kg of ice, but it is the time factor that is confusing me. Is this Q/t=(mL)/t okay, because if I don't put the time on the other side I get something like kg = kg/s
Anyway, I still come up with 3.1045*10^-5 kg/s approximately.
That looks right. It takes about 32 seconds to melt 1 gram.

AM
 
  • #5
Alright, thank you again.
 

1. What is the basic principle behind calculating mass of melting ice in a copper rod?

The basic principle behind calculating the mass of melting ice in a copper rod is the conservation of mass. This means that the mass of the ice and copper rod before melting is equal to the mass of the water and copper rod after melting.

2. How do you determine the mass of the copper rod in the calculation?

The mass of the copper rod can be determined by using a scale or balance to measure its weight. This weight can then be converted to mass using the formula mass = weight / gravitational acceleration.

3. What is the role of temperature in calculating the mass of melting ice in a copper rod?

Temperature plays a crucial role in this calculation as it affects the density and volume of both the ice and the water. The higher the temperature, the lower the density and volume of the water, which will affect the overall mass of the system.

4. How do you account for any heat loss during the melting process?

To account for heat loss during the melting process, we can use a calorimeter to measure the amount of heat released by the melting ice. This heat can then be subtracted from the total amount of heat required to melt the ice, giving a more accurate calculation of the mass of the ice.

5. Can this calculation be applied to other materials besides copper?

Yes, this calculation can be applied to other materials as long as the principle of conservation of mass is followed. However, the specific heat capacity and melting point of the material will need to be taken into consideration when performing the calculation.

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