Water temperature mixing

The mother prepared half a liter of cold water for the children to drink. It mixed water at 20 °C with 400 ml of cold water at 5 °C. What was the temperature of the prepared drink?

Final Answer:

t =  8 °C

Step-by-step explanation:

$$\begin{gathered} V=1/2=\frac{1}{2}=0.5\text{l} \\ {V}_1=400\text{ml}\to \text{l}=400:1000\text{l}=0.4\text{l} \\ {t}_1=5\text{°C} \\ {V}_2=V-V_1=0.5-0.4=\frac{1}{10}=0.1\text{l} \\ {t}_2=20\text{°C} \\ t\cdot V=t_1\cdot V_1+t_2\cdot V_2 \\ t\cdot 0.5=5\cdot 0.4+20\cdot 0.1 \\ 0.5t=4 \\ t=\frac{4}{0.5}=8 \\ t=8=8\text{°C} \end{gathered}$$

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