Question

When 120g of liquid L1 at 65°C was mixed with Yg of liquid L2 at 35°C the final temperature was 45°C. If the specific heat capacity of L2 is half that of L1 find the mass Yg

Answer 1

Let's denote: $m_1 = 120g$ is the mass of L1, $T_1 = 65\degree C$ is the initial temperature of L1, $T_2 = 35\degree C$ is the initial temperature of L2, $T_0 = 45\degree C$ is the final temperature of the mixture, $c_2$ - specific heat capacity of L2, $c_1 = 2c_2$ - specific heat capacity of L1. $m_2 = Yg$ is the mass of L2.

From the energy balance obtain:

$$c_1m_1(T_1 - T_0) = c_2m_2(T_0-T_2)\\ m_2 = \dfrac{2c_2m_1(T_1 - T_0)}{c_2(T_0-T_2)}\\ m_2 = \dfrac{2m_1(T_1 - T_0)}{T_0-T_2}$$

Numerically obtain:

$$m_2 = \dfrac{2\cdot 120(65 - 45)}{45-35} = \dfrac{240\cdot 20}{10} = 480g$$

Answer. 480g.