The bath

Dad poured 50 liters of water at 60 °C into the tub. How much cold water must he add at 10 °C to obtain a bath that has a temperature of 40 °C?

Final Answer:

V₂ =  33.3333 l

Step-by-step explanation:

$$\begin{gathered} V_1=50\text{l} \\ {t}_1=60\text{°C} \\ {t}_2=10\text{°C} \\ t=40\text{°C} \\ Q=m\cdot c\cdot (t-t_0) \\ m=\rho V \\ Q=\rho V\cdot c\cdot (t-t_0) \\ {Q}_1+Q_2=Q \\ \rho \cdot V_1\cdot c\cdot (t_1-t_0)+\rho \cdot V_2\cdot c\cdot (t_2-t_0)=\rho \cdot (V_1+V_2)\cdot c\cdot (t-t_0) \\ {V}_1\cdot (t_1-t_0)+V_2\cdot (t_2-t_0)=(V_1+V_2)\cdot (t-t_0) \\ {V}_1\cdot t_1+V_2\cdot t_2=(V_1+V_2)\cdot t \\ 50\cdot 60+V_2\cdot 10=(50+V_2)\cdot 40 \\ 30V_2=1000 \\ {V}_2=\frac{1000}{30}=33.33333333 \\ {V}_2=\frac{100}{3}\doteq 33.333333=33.3333\text{l} \end{gathered}$$

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