If you have 0.241 m3 of water at 25.0°C and add 0.142 m3 of water at 95.0°C, what is the final temperature of the mixture? Use 1000 kg/m3 as the density of water at any temperature.

Question

matt101 · Answer

This is a thermochemistry problem, so we're talking about a transfer of heat. The heat is being transferred from the hot water (1) to the cold water (2). That means you can set up the following equation:

$$m_1 c \Delta T_1=-m_2 c \Delta T_2$$
We're given volumes, not masses, in the question. However, you can find the mass of each portion of water using the density given in the question (remember, d=m/V, so m=dV). That means we can update the equation:

$$dv_1 c \Delta T_1=-dv_2 c \Delta T_2$$
We can make this equation simpler because density (d) and specific heat of water (c) appear on both sides. That means we can take them out completely:

$$v_1 \Delta T_1=-v_2 \Delta T_2$$
Finally, remember that the change in temperature is final temperature, T(f), minus initial temperature, T(i). The final temperature of both portions of water is going to be the same, and this is what you're solving for. All the other information is given to you in the question! Just plug in the numbers to this equation, solving for T(f):

$$v_1 (T_f-T_{i_1})=-v_2 (T_f-T_{i_2})$$
If you have any questions let me know!