Mixing Problems

The amount of heat lost a hot object can be assumed to be absorbed by a colder object, assuming that the system is well insulated. Mathematically, however, these quantities can only become equal if an extra negative sign is inserted.

- heat lost by hot object = heat gained by cold object

(Use Q = mc DT to obtain the heat for each respective object)

Example 1 What final temperature will be attained if 300. grams of 30.0^o C water are mixed with an equal mass of 66.0 ^o C alcohol? The specific heat of the alcohol is 2.3 J/(g ^o C). Comment on why the final temperature of the mixture is NOT simply the average of the two liquids’ temperature.

Solution

- heat lost by hot object (alcohol = a)t = heat gained by cold object (water = w)

- m_ac_aDT_a = m_wc_w DT_w.

-300g(2.3 J/(g ^o C) )(x – 66) ^o C = 300g(4.19J/(g ^o C) )(x – 30) ^o C

Notice that we use the same final temperature on each side of the equation.

x = 42.8 ^oC

The final temperature is lower than the average temperature, specifically closer to that of water. This is expected because water has a higher specific heat than alcohol.