A) 2800 J/kg-K
B) 2100 J Ag-K
C) 1400 J/kg-K
D) 1200 J/ kg-K
Correct Answer: C
Solution :
Newton's law of cooling is \[mc=\frac{dT}{dt}=\sigma ({{T}^{4}}-T_{0}^{4})A\] \[\Rightarrow \] \[c=\frac{\sigma ({{T}^{4}}-T_{0}^{4})A}{m\left( \frac{dT}{dt} \right)}\] \[\Rightarrow \] \[c=\frac{\begin{align} & (5.73\times {{10}^{-8}})[{{(400)}^{4}}-{{(300)}^{4}}] \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\times 19.2\times {{10}^{-4}} \\ \end{align}}{(34.38\times {{10}^{-3}})\times 0.04}\] \[\Rightarrow \] \[c=1400J/kg-K\]You need to login to perform this action.
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