Finding the rate of heat transfer

[J-Girl]

Hi there, I have this complicated question and I am not sure that I am converting everything corectly. The question is:
"For conduction, the rate of heat transfer is given by $\Delta$Q/$\Delta$t=-kA($\Delta$T/$\Delta$x)

Data given is:

k=806 W.m^-1.K^-1
A=2.53m^2
$\Delta$T=-54.6$^{}$0C
$\Delta$x=5.50x10^-3m

"calculate the rate of energy transfer $\Delta$Q/$\Delta$t, and give its correct S.I unit

I just substituted all of the data into the equation, but didnt know if I had to convert meters^2 into meters?
so this is what I had:
$\Delta$Q/$\Delta$t= -(806)(2.53)[(-54.6)/(5.50x10^-3)]
and my answer was 20243496=2.02 x 10^7 W.m^2.K-1(J.s^-1.m^-2.K^-1)- is there a simple unit name for the answer?

 

[cepheid]

Hi there, I have this complicated question and I am not sure that I am converting everything corectly. The question is:
"For conduction, the rate of heat transfer is given by $\Delta$Q/$\Delta$t=-kA($\Delta$T/$\Delta$x)

Data given is:

k=806 W.m^-1.K^-1
A=2.53m^2
$\Delta$T=-54.6$^{}$0C
$\Delta$x=5.50x10^-3m

"calculate the rate of energy transfer $\Delta$Q/$\Delta$t, and give its correct S.I unit

I just substituted all of the data into the equation, but didnt know if I had to convert meters^2 into meters?
so this is what I had:
$\Delta$Q/$\Delta$t= -(806)(2.53)[(-54.6)/(5.50x10^-3)]
and my answer was 20243496=2.02 x 10^7 W.m^2.K-1(J.s^-1.m^-2.K^-1)- is there a simple unit name for the answer?

You cannot convert m^2 into m, this is nonsensical.

As for the final answer, you should end up with something that has dimensions of power (i.e. energy/time). This is clear from looking at the left-hand side: delta Q/delta t. Therefore, your answer should just have units of watts, and if it doesn't, then you've made a mistake.

 

[J-Girl]

oh okay thanks:) but in the working out, is everything right? I mean, converted correctly to get a number of 2.02 x 10^7 ?

 

[Sri_Govind]

Let us look at the dimensions of the right hand side of your equation

the thermal conductivity K has the units of W/mK

A is the area available for heat transfer and thus has a unit of m^2

dT/dX has the unit of K/m

so in all, the RHS has the units of W (Joules/Second)

the LHS must also have the same dimension !

 

[Sri_Govind]

As a followup, your substitutions seem to be OK, just that your units do not match up

 

[cepheid]

Sri_Govind: this thread is 10 months old, and it is highly unlikely that the OP will check it any longer. Resurrecting dead threads for no reason (called "necro-posting") is strongly frowned upon. Don't do it.