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

$$mc(T_1-T)=Mc(T-T_2)\\\rho V_1c(T_1-T)=\rho V_2c(T-T_2)\\ 80(67-T)=300(T-0)\\T=14.1\degree C$$