Water mixing

We have 520 ml of hot water and 640 ml of water at 48°C. What is the temperature of approximately hot water when the resulting mixture has a temperature of 65°C?

Correct answer:

t₁ =  85.9231 °C

Step-by-step explanation:

$$\begin{gathered} V_1=520\text{ml} \\ {V}_2=640\text{ml} \\ {t}_2=48{}^{\circ } \\ t=65{}^{\circ } \\ Q=m\cdot c\cdot (t-t_0) \\ m=\rho V \\ Q=\rho V\cdot c\cdot (t-t_0) \\ {Q}_1+Q_2=Q \\ \rho \cdot V_1\cdot c\cdot (t_1-t_0)+\rho \cdot V_2\cdot c\cdot (t_2-t_0)=\rho \cdot (V_1+V_2)\cdot c\cdot (t-t_0) \\ {V}_1\cdot (t_1-t_0)+V_2\cdot (t_2-t_0)=(V_1+V_2)\cdot (t-t_0) \\ {V}_1\cdot t_1+V_2\cdot t_2=t\cdot (V_1+V_2) \\ 520\cdot t_1+640\cdot 48=65\cdot (520+640) \\ 520t_1=44680 \\ {t}_1=\frac{44680}{520}=85.92307692 \\ {t}_1=\frac{1117}{13}\doteq 85.923077=85.9231\text{°C} \end{gathered}$$

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