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

Click For Summary
SUMMARY

The discussion centers on calculating the mass of ice that melts per second in a copper rod scenario, where one end is in boiling water and the other in an ice-water mixture. The thermal conductivity of copper (k = 390) and the dimensions of the rod (length = 1.5 m, cross-sectional area = 4.00 x 10^-4 m^2) are critical to the calculations. The heat transfer equation Q = (k * A * ΔT * t) / L is applied, with ΔT assumed to be 100°C. The final calculation yields a mass flow rate of approximately 3.1045 x 10^-5 kg/s for the melting ice.

PREREQUISITES
  • Understanding of thermal conductivity and its role in heat transfer.
  • Familiarity with the concept of latent heat, specifically for ice melting (L = 33.5 x 10^4 J/kg).
  • Basic knowledge of heat transfer equations, particularly Q = (k * A * ΔT * t) / L.
  • Ability to manipulate units and equations to solve for mass flow rates.
NEXT STEPS
  • Study the principles of heat transfer in solids, focusing on Fourier's law.
  • Learn about the latent heat of fusion and its applications in phase change problems.
  • Explore practical examples of thermal conductivity in different materials.
  • Practice solving similar physics problems involving heat transfer and phase changes.
USEFUL FOR

This discussion is beneficial for high school physics students, educators teaching thermal dynamics, and anyone interested in practical applications of heat transfer principles.

2099
Messages
4
Reaction score
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
 
Physics news on Phys.org
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
 
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.
 
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
 
Alright, thank you again.
 

Similar threads

Ice cube in water: entropy calculation (Calorimetry)
  • · Replies 6 ·
Replies
6
Views
670
Confusion about whether to use the specific heat of water or ice
  • · Replies 4 ·
Replies
4
Views
2K
Using thermal conductivity to melt ice
  • · Replies 4 ·
Replies
4
Views
2K
Thermal conduction between 3 rods
  • · Replies 2 ·
Replies
2
Views
10K
Steady state heat flow: radiation and conduction
  • · Replies 5 ·
Replies
5
Views
3K
Thermal Conduction Homework: Find Rod's Conductivity
  • · Replies 7 ·
Replies
7
Views
5K
Finding thermal conductivity of an insulated metal bar
  • · Replies 2 ·
Replies
2
Views
15K
Thermodynamics Ice Melting Question
  • · Replies 4 ·
Replies
4
Views
3K
Conducting rod in equilibrium due to magnetic force
  • · Replies 13 ·
Replies
13
Views
2K
Modeling a kongming lantern (sky lantern)
  • · Replies 6 ·
Replies
6
Views
3K