- #1
Brianjw
- 40
- 0
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:
[tex] .0078kg * 3.34 * 10^5 J/Kg = 2605.2 [/tex]
then I use the formula:
[tex] H = Area*k*(T_h - T_c)/L [/tex]
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 forumla and converting Celcius to Kelvin I have:
[tex] 230.549J/min = .000106 m^2 * k * 373.15K/.733m^2 [/tex]
which gives me approx 4270. Can anyone see where I went wrong?
Thanks
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:
[tex] .0078kg * 3.34 * 10^5 J/Kg = 2605.2 [/tex]
then I use the formula:
[tex] H = Area*k*(T_h - T_c)/L [/tex]
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 forumla and converting Celcius to Kelvin I have:
[tex] 230.549J/min = .000106 m^2 * k * 373.15K/.733m^2 [/tex]
which gives me approx 4270. Can anyone see where I went wrong?
Thanks