In Newton's experiment of cooling, the water equivalent of two similar calorimeters is $10 $ gm each. They are filled with $350 gm$ of water and $300 gm$ of a liquid (equal volumes) separately. The time taken by water and liquid to cool from ${70^o}C$ to ${60^o}C$ is $3$ min and $95$ sec respectively. The specific heat of the liquid will be ...... $Cal/gm\,^oC$
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(c) ${S_l} = \frac{1}{{{m_l}}}\left[ {\frac{{{t_l}}}{{{t_W}}}({m_W}{C_W} + W) - W} \right]$
$ = \frac{1}{{300}}\left[ {\frac{{95}}{{3 \times 60}}(350 \times 1 + 10) - 10} \right]$ $=0.6 Cal/gm\times°C$
$ = \frac{1}{{300}}\left[ {\frac{{95}}{{3 \times 60}}(350 \times 1 + 10) - 10} \right]$ $=0.6 Cal/gm\times°C$
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