Question

perfectly insulated container with no heat capacity is used to mix together the following three samples of water:

I.183.2 grams of water at 23.2 °C

II.32.1 grams of water at 14.2 °C

III.143 grams of water at an unknown temperature.

After the three samples are combined, the final temperature of the combined solutions is 39.0 °C. What is the temperature of sample III before the samples were mixed together?

Answer 1

Ans.     q = m C x dT    - equation 1

            where, q = heat content

                        m = mass

C = heat capacity of calorimeter system = 4.184 J/⁰C

dT = final – initial temperature

Let the temperature of water sample be T.

Now, heat content water sample 1, q1 = 183.2 g x (4.184 J/⁰C) x 23.2⁰C

                                                            = 17629.7024 J

Heat content, q2 = 32.1 g x (4.184 J/⁰C) x 14.2⁰C = 1907.15088 J

Heat content, q3 = 143.0 g x (4.184 J/⁰C) x T⁰C = 598.312 T⁰C

Total heat content of the three-water sample =

            q1 + q2 + q3 = 12110.83904 J + 3330.79872 J + 23334.168 J - 598.312 T

            or, q (total) = 29995.76544 - 598.312 T

Total mass of water after mixing = 183.2 g + 32.1 g + 143.0 g = 358.3 g

Heat content of the mixture = 358.3 g x (4.184 J/⁰C) x 39.0 ⁰C = 58465.9608 J

Now,

Total heat content of the mixture = heat content of (q1 + q2 + q3)

Or, 58465.9608 J = 17629.7024 J + 1907.15088 J + 598.312 T⁰C

Or, 598.312 T⁰C = 38929.10752 J

Or, T⁰C = 38929.10752 J / 598.312 = 65.06

Thus, the initial temperature of water III = 65.06⁰C