Answer to Question #124953 in Molecular Physics | Thermodynamics for KUMAH EMMANUEL

Question

A hot liquid at 80 ̊ C is added to 600 g of the same liquid originally at 10 ̊ C. When the
mixture reaches 30 ̊ C, what will be the total mass of the liquid?

Answer 1

$T_1 = 80^\circ C$

$T_2 = 10^\circ C$

$T_0 = 30^\circ C$

$M_2 = 600 g$

$M_1 +M_2 = ?$

$с_1M_1(T_0-T_1) +c_2M_2(T_0-T_2) = 0$ $c_1=c_2$ (the same liquid)

Then $M_2(T_0 - T_2) = M_1(T_1 - T_0)$ $\Rightarrow M_1 = \dfrac{M_2(T_0 - T_2)}{T_1 - T_0} = 600g \cdot \dfrac{30 - 10}{80 - 30} = 240g$

Finally, $M_1 +M_2 = 240g + 600g = 840g$