Bravo! That is correct.You know the equation for the heat-transfer rate,

H _{cd}= Q/t = K A (T_{2}- T_{1}) / L(Remember, lower case t is time and upper case T is temperature). We have all the pieces for the terms on the right so we can just do the calculation.

While it is easier to say or to think of the area in terms of square-centimeters, we will need to convert that to square meters,

A = 2.2 cm ^{2}( 1 m / 100 cm )^{2}A = 2.2 x 10

^{ - 4}m^{2}Notice that we have to

squarethe conversion factor, ( 1 m / 100 cm ). That's important! Now we need to make a similar conversion with the length. Again, while it is easier to say "thirty centimeters" or to think of "thirty centimeters" for the length, by the time we get ready for calculations, we need the length in^{2}meters,L = 0.30 m H

_{cd}= Q/t = K A (T_{2}- T_{1}) / LH

_{cd}= [240 W/m C^{o}][2.2 x 10^{ - 4}m^{2}][1800^{o}C - 20^{o}C] / 0.30 mH

_{cd}= [240 W/m C^{o}][2.2 x 10^{ - 4}m^{2}][1780 C^{o}] / 0.30 mH

_{cd}= [(240)(2.2 x 10^{ - 4})(1780 )/0.30][(W/m C^{o})( m^{2})(C^{o})/m]H

_{cd}= [(240)(2.2 x 10^{ - 4})(1780 )/0.30][W]

H_{cd}= 313 W

(c) 2000, Doug Davis; all rights reserved.