Question

A metal ball of mass $0.1kg$ is heated upto $500^{\circ }C$ and dropped into a vessel of heat capacity $800J{K}^{-1}$ and containing $0.5kg$ water. The initial temperature of water and vessel is $30^{\circ }C$. What is the approximate percentage increment in the temperature of the water ?
[Specific Heat Capacities of water and metal are, respectively, $4200Jk{g}^{-1}{K}^{-1}$ and $400JK{g}^{-1}{K}^{-1}$]

• A. 30 %
• B. 20 %
• C. 25 %
• D. 15 %

Answer 1

Answer: (B)

$0.1\times 400\times (500-T)=0.5\times 4200\times (T-30)+800(T-30)$
$\Rightarrow 40(500-T)=(T-30)(2100+800)$
$\Rightarrow 20000-40T=2900T-30\times 2900$
$\Rightarrow 20000+30\times 2900=T(2940)$
$T=30.4^{\circ }C$
$\frac{\Delta T}{T}\times 100=\frac{6.4}{30}\times 100$
$\simeq$ 20 %