Question

Suppose you mix 7.00 L of water at \(2.00 \cdot 10^{1}{ }^{\circ} \mathrm{C}\) with \(3.00 \mathrm{~L}\) of water at \(32.0^{\circ} \mathrm{C}\); the water is insulated so that no energy can flow into it or out of it. (You can achieve this, approximately, by mixing the two fluids in a foam cooler of the kind used to keep drinks cool for picnics.) The \(10.0 \mathrm{~L}\) of water will come to some final temperature. What is this final temperature?

Short answer

Answer: The final temperature of the mixed water is 23.6°C.

Step-by-step solution

Convert volumes to masses

Since we know the volume of water and the density of water is \(1\, \text{g/mL}\), we can convert the volume of water into mass for further calculations. The mass of the fluids can be calculated by multiplying their volume by the density. For water at \(20^{\circ} \mathrm{C}\): \(mass_{1} = density \times volume_{1}\) \(mass_{1} = 7.00\, \mathrm{L} \times (1\, \text{g/mL} \times 1000\, \text{mL/L})\) \(mass_{1} = 7,000\, \text{g}\) For water at \(32^{\circ} \mathrm{C}\): \(mass_{2} = density \times volume_{2}\) \(mass_{2} = 3.00\, \mathrm{L} \times (1\, \text{g/mL} \times 1000\, \text{mL/L})\) \(mass_{2} = 3,000\, \text{g}\)

Set up the heat transfer equations

We know that the heat transfer equation is \(q = mc\Delta T\). The heat gained by the colder water is equal to the heat lost by the warmer water. The specific heat capacity of water is \(4.18\, \text{J/(g}\,^{\circ} \mathrm{C)}\). \(mass_{1}c\Delta T_{1} = mass_{2}c\Delta T_{2}\) We can also write the temperature difference as: \(\Delta T_{1} = T_{final} - T_{1}\) \(\Delta T_{2} = T_{2} - T_{final}\) Substitute these into the heat transfer equation: \(mass_{1}c(T_{final} - T_{1}) = mass_{2}c(T_{2} - T_{final})\)

Solve the equation for the final temperature

Now, we can calculate the final temperature by solving the equation: \(mass_{1}c(T_{final} - T_{1}) = mass_{2}c(T_{2} - T_{final})\) \(7,000\, \text{g} \times 4.18\, \mathrm{J/(g}\,^{\circ} \mathrm{C)}(T_{final} - 20^{\circ} \mathrm{C}) = 3,000\, \text{g} \times 4.18\, \mathrm{J/(g}\,^{\circ} \mathrm{C)}(32^{\circ} \mathrm{C} - T_{final})\) \(c\) can be removed from both sides: \(7,000\, \text{g} (T_{final} - 20^{\circ} \mathrm{C}) = 3,000\, \text{g} (32^{\circ} \mathrm{C} - T_{final})\) Expanding the equation gives: \(7,000\, \text{g} \times T_{final} - 140,000\, \mathrm{g^{\circ} \mathrm{C}} = 96,000\, \mathrm{g^{\circ} \mathrm{C}} - 3,000\, \mathrm{g} \times T_{final}\) Combine terms and divide by the total mass to find the final temperature: \(T_{final} = \frac{140,000\, \mathrm{g^{\circ} \mathrm{C}} + 96,000\, \mathrm{g^{\circ} \mathrm{C}}}{7,000\, \text{g} + 3,000\, \text{g}}\) \(T_{final} = \frac{236,000\, \mathrm{g^{\circ} \mathrm{C}}}{10,000\, \text{g}}\) \(T_{final} = 23.6^{\circ}\mathrm{C}\) The final temperature of the mixed water is \(23.6^{\circ}\mathrm{C}\).