Question #93a6f

Oct 6, 2016

Given
${m}_{w} \to \text{Mass of water"="its volume"xx} \mathrm{de} n s i t y$
$= 25 c {m}^{3} \times 1 \frac{g}{c {m}^{3}} = 25 g$

${t}_{w} \to \text{Initial temperature of water} = {23}^{\circ} C$

${m}_{C d} \to \text{Mass Cadmium} = 65.5 g$

${t}_{C d} \to \text{Initial temperature of Cd} = {100}^{\circ} C$

${C}_{w} \to \text{Sp.heat capacity of water} = 4.184 \frac{J}{{g}^{\circ} C}$

${C}_{C d} \to \text{Sp.heat capacity of Cd} = 0.2311 \frac{J}{{g}^{\circ} C}$

Let the final temperature of the system be ${t}^{\circ} C$

So by calorimetric principle

$\text{Heat lost by Cd" = "Heat gained by water}$

$\implies {m}_{C d} \times {C}_{C d} \times \left({t}_{C d} - t\right) = {m}_{w} \times {C}_{w} \times \left(t - {t}_{w}\right)$

$\implies 65.5 \times 0.2311 \times \left(100 - t\right) = 25 \times 4.184 \times \left(t - 23\right)$

$\implies \frac{65.5 \times 0.2311}{25 \times 4.184} \times \left(100 - t\right) = \left(t - 23\right)$

$\implies 0.1447 \times \left(100 - t\right) = \left(t - 23\right)$

$\implies 14.47 - 0.1447 t = \left(t - 23\right)$

$\implies 1.1447 t = 37.47$

$\implies t = \frac{37.47}{1.1447} \approx {32.73}^{\circ} C$