Thermal Question, I seem to be doing something wrong

[Brianjw]

Well I swear I am doing this right but something seems that its the wrong answer as the website I enter it into is rejecting it. If you see a flaw in my methods let me know:

The two ends of an insulated metal rod are maintained at a temperature differential of 100 degrees C . The rod has a length of 73.3 cm and a cross-sectional area of 1.06 cm^2. The heat conducted by the rod melts a mass of 7.80g of ice in a time of 11.3 min

So what I did was first find the amount of energy required to melt the ice without changing the temperature which uses Q = m*L

therefore:

.0078kg * 3.34 * 10^5 J/Kg = 2605.2

then I use the formula:

H = Area*k*(T_h - T_c)/L

I need to find k for the answer in terms of W/m*k

So since H = dQ/dt I get, 2605.2/11.3 = 230.549
Using the above formula and converting celsius to Kelvin I have:

230.549J/min = .000106 m^2 * k * 373.15K/.733m^2

which gives me approx 4270. Can anyone see where I went wrong?

Thanks

 

[Doc Al]

Using the above formula and converting celsius to Kelvin I have:

230.549J/min = .000106 m^2 * k * 373.15K/.733m^2

For some reason you treated the temperature difference as if it were a temperature that needed converting. No need to "convert": 100 C-degrees = 100 K-degrees.

Another problem: measure heat flow in J/sec, not J/min.

 

[M98Ranger]

for sharing your approach and calculations. It seems like your method and calculations are correct, but there may be a few potential areas where things could have gone wrong. Here are a few things to consider:

1. Make sure you are using the correct units throughout your calculations. In your first equation, you used J/kg as the unit for specific latent heat, but in the second equation, you used J/min as the unit for heat transfer rate. Make sure to convert all units to the same system before plugging them into equations.

2. Double check your cross-sectional area. In the second equation, you used 0.000106 m^2, but in the given information, the cross-sectional area is 1.06 cm^2. Make sure to convert the units to the same system before using them in calculations.

3. Check your conversion from Celsius to Kelvin. In the second equation, you used 373.15 K, but that is the conversion from Celsius to Fahrenheit, not Kelvin. The correct conversion is simply adding 273.15 to the Celsius temperature.

4. Lastly, make sure to check your final answer and see if it makes sense. In this case, the value of 4270 for thermal conductivity seems a bit high. It may be worth double checking your calculations to see if there was a mistake somewhere.

I hope this helps! Good luck with your calculations.