Question

During the summer after your first year at Carnegie Mellon, you are lucky enough to get a job making coffee at Starbucks, but you tell your parents and friends that you have secured a lucrative position as a "java engineer." An eccentric chemistry professor (not mentioning any names) stops in every day and orders 200ml of Sumatran coffee at precisely 60.0°C. You then need to add enough milk at 1.00°C to drop the temperature of the coffee, initially at 80.0°C, to the ordered temperature.
Calculate the amount of milk (in ml) you must add to reach this temperature. Show all your work in the provided spaces.

In order to simplify the calculations, you will start by assuming that milk and coffee have the specific heat and density as if water. In the following parts, you will remove these simplifications. Solve now this problem assuming the density is 1.000 g/ml for milk and coffee and their specific heat capacity is 4.184 J/(g ºC).

Hint: the coffee is in an insulated travel mug, so no heat escapes. To insulate a piece of glassware in Virtual Lab, Mac-users should command-click (or open-apple click) on the beaker or flask; Windows users should right click on the beaker or flask. From the menu that appears choose “Thermal Properties.” Check the box labeled “insulated from surroundings.” The temperature of the solution in that beaker or flask will remain constant.

Answer 1

Answer:

67.80 mL

Explanation:

The heat q is given by the formula:

$q = mC_p(T_2-T_1)$

Since heat released by the coffee =  - heat absorbed by the milk; Then :

$q_{coffee}= -q_{milk}$

$m_{coffee}C_p(T_2-T_1)_{coffee} = -m_{milk}C_p(T_2-T_1)_{milk}$

From the question;

given that the heat capacity of milk & coffee are equal; The density are also said to be equal:

The temperature difference for coffee is as follows:

$(T_2-T_1)_{coffee} = 60.00^ ^0 }}C - 80.00 ^ ^ 0 }}C = -20.00 ^ ^ 0 }}C$

Temperature difference for milk is :

$(T_2-T_1)_{milk} = 60.00^ ^0 }}C - 1.00 ^ ^ 0 }}C = 59.00 ^ ^ 0 }}C$

We all know that : Density $\rho$ = mass (m) / volume (v)

then m = v × $\rho$

So, we can say that : $m_{coffee}C_p(T_2-T_1)_{coffee} = -m_{milk}C_p(T_2-T_1)_{milk}$ can now be re-written as:

$v_{coffee}*\rho _{coffee}* (T_2-T_1)_{coffee} = -v_{milk}*\rho_{milk}*(T_2-T_1)_{milk}$

$v_{coffee}* (T_2-T_1)_{coffee} = -v_{milk}*(T_2-T_1)_{milk}$

Replacing the values; we have :

$200 \ ml * (-20.0^0 C)= -v_{milk}*(59^0 C)$

$v_{milk} = \frac{200 \ ml *(-20.0^0 C)}{-59^0 \ C}$

$v_{milk} = 67.80 \ mL$

Therefore, the amount of milk required to reach the required temperature for coffee is 67.80 mL