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A metal bal of mass $0.1\, kg$ is heated upto $500\,{}^oC$ and dropped into a vessel of heat capacity $800\, JK^{-1}$ and containing $0.5\, kg$ water. The initial temperature of water and vessel is $30\,{}^oC$. ........ $\%$ is the approximate percentage increment in the temperature of the water. [Specific heat Capacities of water and metal are, respectively $4200\, Jkg^{-1}K^{-1}$ and $400\, Jkg^{-1}K^{-1}$]
$15$
$30$
$25$
$20$
Solution
$0.1 \times 400 \times \left( {500 – T} \right)$
$ = 0.5 \times 4200 \times \left( {T – 30} \right) + 800\left( {T – 30} \right)$
$ \Rightarrow 40\left( {500 – T} \right) = \left( {T – 30} \right)\left( {2100 + 800} \right)$
$ \Rightarrow 20000 – 40T = 2900\,T – 30 \times 2900$
$ \Rightarrow 20000 + 30 \times 2900 = T\left( {2940} \right)$
$T = {30.4^ \circ }C$
$\frac{{\Delta T}}{T} \times 100 = \frac{{6.4}}{{30}} \times 100 \simeq 20\% $
