A) 24 cal
B) 56 cal
C) 82 cal
D) 100 cal
Correct Answer: C
Solution :
[c] Heat required to raise the temperature of m grams of substance by dT is given as \[dQ=mc\,dT\Rightarrow Q=\int{mcdT}\] Therefore, to raise the temperature of 2g of substance from \[5{}^\circ C\]to \[15{}^\circ C\] \[Q=\int\limits_{5}^{15}{2\times (0.2+0.14t+0.023{{t}^{2}})dT}\] \[=2\times {{\left[ 0.2t+\frac{0.14{{t}^{2}}}{2}+\frac{0.023{{t}^{3}}}{3} \right]}_{5}}^{15}=82cal\]You need to login to perform this action.
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