Question

what volume of water in ml initially at 85.4 C needs to be mixed with 200 ml of water initally at 29.5 C so that the final temperature of the water is 36.2 C

Answer 1

Answer:

About 27 mL of water.

Explanation:

We can use the heat transfer equation. Recall that:

$\displaystyle q = mC\Delta T$

Heat is transferred from the water with higher temperature to the water with lower temperature. Hence:

$\displaystyle -q_1 = q_2$

Substituting yields:

$-m_1C \Delta T_1 = m_2C\Delta T _2$

We can solve for m₁:

$\displaystyle \begin{aligned} m_1 &= -\frac{m_2 C\Delta T_2}{ C \Delta T_1} \\ \\ & = -\frac{m_2\Delta T_2}{\Delta T _1}\end{aligned}$

The desired final temperature of water is 36.2 °C. Substitute and evaluate:
$\displaystyle \begin{aligned} m_1 & = -\frac{(200\text{ mL})(36.2\text{ ^\circ C}-29.5\text{ ^\circ C})}{(36.2\text{ ^\circ C}-84.5\text{ ^\circ C})} \\ \\ & = -\frac{(200\text{ mL})(6.7\text{ ^\circ C})}{-49.2\text{ ^\circ C}}\\ \\ & =27\text{ mL}\end{aligned}$

In conclusion, about 27 mL of water should be added.