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Q. A metal ball of mass $0.1\, kg$ is heated upto $500^{\circ}C$ 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 $30^{\circ}C$. What 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 $400 \; JKg^{-1} K^{-1}$]

JEE MainJEE Main 2019Thermal Properties of Matter

Solution:

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