Question

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 $400 \; JKg^{-1} K^{-1}$]

• A. 30%
• B. 20%
• C. 25%
• D. 15%

Answer 1

Answer: (B)

20%

Answer 2

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