A) 0.3 Cal/gm ´°C
B) 0.5 Cal/gm ´°C
C) 0.6 Cal/gm ´°C
D) 0.8 Cal/gm ´°C
Correct Answer: C
Solution :
\[{{S}_{l}}=\frac{1}{{{m}_{l}}}\left[ \frac{{{t}_{l}}}{{{t}_{W}}}({{m}_{W}}{{C}_{W}}+W)-W \right]\] \[=\frac{1}{300}\left[ \frac{95}{3\times 60}(350\times 1+10)-10 \right]\]=0.6 Cal/gm´°CYou need to login to perform this action.
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