A) 4.8 cal/s
B) 6.0 cal/s
C) 1.2 cal/s
D) 2.4 cal/s
Correct Answer: A
Solution :
\[H={{H}_{1}}+{{H}_{2}}\] \[\frac{100-T}{\frac{{{\ell }_{1}}}{{{k}_{1}}A}}=\frac{T-0}{\frac{{{\ell }_{2}}}{{{k}_{2}}A}}+\frac{T-0}{\frac{{{\ell }_{3}}}{{{k}_{3}}A}}\] \[\frac{{{k}_{1}}(100-T)}{{{\ell }_{1}}}=\frac{{{k}_{2}}T}{{{\ell }_{2}}}+\frac{{{k}_{3}}T}{{{\ell }_{3}}}\] \[\frac{0.92}{46}(100-T)=\frac{0.26}{13}T+\frac{0.12}{12}T\] \[2\times {{10}^{-2}}(100-T)=2\times {{10}^{-2}}T+1\times {{10}^{-2}}T\] \[2(100-T)=2T+T\] \[2(100-T)=3T\] \[200=5T\] \[T={{40}^{o}}C\] \[H=\frac{{{k}_{1}}(100-T)A}{{{\ell }_{1}}}=\frac{0.92}{46}(100-10)\times 4\] \[=2\times {{10}^{-2}}\times 60\times 4=120\times {{10}^{-2}}\times 4=4.8\]cal/sYou need to login to perform this action.
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