A) \[{{K}_{1}}{{A}_{1}}={{K}_{2}}{{A}_{2}}\]
B) \[\frac{{{K}_{1}}\,{{A}_{1}}}{{{s}_{1}}},\frac{{{K}_{2}}{{A}_{2}}}{{{s}_{2}}}\]
C) \[{{K}_{2}}{{A}_{1}}={{K}_{1}}{{A}_{2}}\]
D) \[\frac{{{K}_{2}}{{A}_{1}}}{{{s}_{2}}}=\frac{{{K}_{1}}{{A}_{2}}}{{{s}_{1}}}\]
Correct Answer: A
Solution :
| Rate of loss of heat by conduction is. |
| \[H=\frac{\Delta Q}{\Delta t}=K\,\,A\left( \frac{{{T}_{1}}-{{T}_{2}}}{l} \right)\] |
| For first rod. |
| \[{{H}_{1}}={{K}_{1}}\,\,{{A}_{1}}\left( \frac{{{T}_{1}}-{{T}_{2}}}{{{l}_{1}}} \right)\] |
| For second rod, |
| \[{{H}_{2}}={{K}_{2}}{{A}_{2}}\left( \frac{{{T}_{1}}-{{T}_{2}}}{{{l}_{2}}} \right)\] |
| but \[{{l}_{1}}={{l}_{2}}\] and \[{{H}_{2}}={{H}_{2}}\] |
| So, we have |
| \[{{K}_{1}}{{A}_{1}}\,({{T}_{1}}-{{T}_{2}})={{K}_{2}}{{A}_{2}}\,({{T}_{1}}-{{T}_{2}})\] |
| or \[{{K}_{1}}{{A}_{1}}={{K}_{2}}{{A}_{2}}\] |
| Note: Heat transfer occurs only between regions that are at different temperatures, and the rate of heat flow is \[\frac{dQ}{dt}\] |
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