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

About 27 mL of water.

Step-by-step 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 &= -(m_2 C\Delta T_2)/( C \Delta T_1) \\ \\ & = -(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.