Answer to Question #107239 in Chemistry for Pedro Serrano

Question

They took two 100mL water samples at two different temperatures ,mixed them together, and recorded the temperatures. The temperature of the first was 30 degrees and the temperature of the second was 90 degrees. What was the final temperature of the two when mixed together

Answer 1

The quantity of heat $Q$ transferred in this process of mixing is:

$Q = cm\Delta T$ ,

where $c$ is the heat capacity of water, $m$ is the mass of water and $\Delta T$ is the change in temperature (final temperature $x$ minus starting temperature). The quantity of heat given by the sample of water at 90°C is equal to the quantity of heat absorbed by the sample of water at 30°C. As the volumes of the water samples are equal, if one ignores the difference in density of the two water samples (and also the dependence of heat capacity on temperature), the change in temperature of the first sample is equal to the change in temperature of the second sample, multiplied by -1:

$x - 30 = -1\cdot (x-90)$

As you could see, we are not obliged to convert from celsius to kelvin here, as we calculate difference temperature.

The final temperature of the mixture is:

$x = \frac{90+30}{2} = 60$ °C

If one doesn't ignore the difference in density (see the curve here : https://www.engineeringtoolbox.com/water-density-specific-weight-d_595.html), then the products of density and change in temperature are equal:

$d_{30} (x-30) = -d_{90}(x-90)$

$0.99567\cdot(x-30) = -0.9653\cdot(x-90)$

Then, the final temperature is: 59.54 °C

Answer: The final temperature of the two when mixed together is 60°C or, more exactly, 59.54 °C.