Question

80 L of water at a temperature of 67 C is added to a bath containing 300 L of water at 0 C. What is the final temperature of the bath? Take the specific heat capacity of water to be 4200 J/kg/K.

Answer 1

$$Q_1=Q_2\\m_1c(T_1-T)=m_2c(T-T_2)\\\rho V_1(T_1-T)=\rho V_2(T-T_2)\\80(67-T)=300(T-0)\\T=14.1\degree C$$