Question

You drop 8.55 g if aluminum originally at 95*C into 25.1g of water originally at 12.3*C, what is the final temperature of the mixture

Answer 1

$\begin{aligned} m_{al} &= 8.55g\\ m_{water} &= 25.1g\\ T_{al} &= 95°C\\ T_{water} &= 12.3°C\\ c_{al} &= 0.89J/g°C\\ c_{water} &= 4.18J/g°C \end{aligned}$

From heat lost by Al = heat gained by water,

$m_{water}× c_{water}× (T_f - T_i) = m_{al} × c_{al} × (T_i - T_f)$

$25.1 × 4.18 × (T_f - 12.3) = 8.55 × 0.89 × (95 - T_f)$

$104.918 (T_f - 12.3) = 7.6099 (95 - T_f)$

$104.918 T_f - 1290.4914= 722.9405 - 7.6099T_f$

$112.5279T_f = 2013.4319$

$T_f = \dfrac{2013.4319}{112.5279}$

$T_f = 16.43°C$