When you take a bath, how many kg of hot water (49 C) must you mix with cold water (13 C) so that the temperature of the bath is 36 C? The total mass of the water (hot and cold) is 191 kg.

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- Nov 11th 2008, 06:10 AM #1

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- Nov 11th 2008, 06:50 AM #2
When mixing hot water with cold, the resultant solution will reach its equilibrium temperature where the hot water loses heat and the cold water takes in heat. So:

$\displaystyle \begin{aligned} Q_{\text{cold}} & = -Q_{\text{hot}}

\\ (m c \Delta T)_{\text{cold}} & = - (mc\Delta T)_{\text{hot}}

\\ m_{\text{cold}} \cdot 4.184 \frac{J}{g ^{\circ}C} \cdot \left(36 ^{\circ}C - 13 ^{\circ} C\right) & = - \bigg( m_{\text{hot}} \cdot 4.184 \frac{J}{g ^{\circ}C} \cdot \left(36 ^{\circ} C - 49 ^{\circ} C\right) \bigg)

\end{aligned}$

Now, we have two variables to deal with and we only want to solve for $\displaystyle m_{\text{hot}}$. Remember, what they gave you? The total mass of both hot and cold water is 191 kg, i.e. $\displaystyle m_{\text{cold}} + m_{\text{hot}} = 191$ which means: $\displaystyle m_{\text{cold}} = 191 - m_{\text{hot}}$

Substitute that in and you should be able to solve for $\displaystyle m_{\text{hot}}$