A) \[10{}^\circ C/cm\]
B) \[12{}^\circ C/cm\]
C) \[25{}^\circ C/cm\]
D) \[20{}^\circ C/cm\]
Correct Answer: C
Solution :
\[\frac{\Delta Q}{\Delta t}=\frac{KA\,\Delta \theta }{\Delta x}\]Þ Thermal gradient \[\frac{\Delta \theta }{\Delta x}\] \[=\frac{(\Delta Q/\Delta t)}{K}=\frac{10}{0.4}=25{}^\circ C/cm\]
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