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A mass of $50\,g$ of water in a closed vessel, with surroundings at a constant temperature takes $2\, minutes$ to cool from $30\,^oC$ to $25\,^oC$. A mass of $100\,g$ of another liquid in an identical vessel with identical surroundings takes the same time to cool from $30\,^oC$ to $25\,^o C$. The specific heat of the liquid is .......... $kcal/kg$ (The water equivalent of the vessel is $30\,g$.)
$2.0$
$7$
$3$
$0.5$
Solution
As the surrounding is identical, vessel is identical time taken to cool both water and liquid$\left( {from\,{{30}^ \circ }C\,to\,{{25}^ \circ }C} \right)$ is same $2\,minutes,$ therefore
${\left( {\frac{{dQ}}{{dt}}} \right)_{water}} = {\left( {\frac{{dQ}}{{dt}}} \right)_{liquid}}$
$or,\frac{{\left( {{m_w}{C_w} + W} \right)\Delta T}}{t} = \frac{{\left( {{m_\ell }{C_\ell } + W} \right)\Delta T}}{t}$
$\left( {W = water\,equivalent\,of\,the\,vessel} \right)$
$or,\,\,{m_w}{C_w} = {m_\ell }{C_\ell }$
$\therefore Specific\,heat\,of\,liquid,\,{C_\ell } = \frac{{{m_W}{C_W}}}{{{m_\ell }}}$
$ = \frac{{50 \times 1}}{{100}} = 0.5\,kcal/kg$
