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 250ml of Sumatran coffee at precisely 75.0°C. You then need to add enough milk at 7.00°C to drop the temperature of the coffee, initially at 85.0°C, to the ordered temperature. Required:Calculate the amount of milk (in ml) you must add to reach this temperature.

Answer 1

Answer:

The answer is "36.76  ml "

Explanation:

In the given question, Its basic understanding to also be hired seems to be that coffee heat = heat that the milk absorbs  

The following relationship defines heat:

$\to q = m C_{p}( T_2 - T_1)$

Its updated equation is:

$\to q_{\ coffee} = - q_\ milk} \\\\ \to m_{\ milk} C_{p} (T_2 - T_1)_{\ coffee} = - m_{ \ milk} C_{p} (T_2 - T_1)_{\ milk} ..........(a)$

The following information is provided to you as per the question:  

1) Milk and coffee have the same heat power  

2) Difference in coffee temperature is:

$\to (T_2 - T_1)_{\ coffee} = 75.0^{\circ} \ \ C-85.0^{\circ} \ \ C$

                              $= -10.0 ^{\circ} \ \ C$

3) The difference in milk temperature:

$\to (T_2 - T_1)_{ \ milk} = 75.0^{\circ} \ \ C- 7.0^{\circ} \ \ C$

$= 68.0 ^{\circ} \ \ C$

4)  Milk and coffee density are equal therefore, weight and volume are equal and have same ratio, that can be shown as follows:

$\to \ Density (\rho) =\frac{ \ Mass (m)}{ \ Volume (V)} \\\\\to m= V \times \rho$

$\to m_{\ coffee} \times (T_2 - T_1)_{ \ coffee}= -m_{\ milk} \times (T_2 - T_1)_{milk} \\\\\to V_{\ coffee} \times \rho_{\ coffee} \times (T_2 - T_1)_{\ coffee} = - V_{\ milk} \times \rho_{\ milk} \times (T_2 - T_1)_{\ milk} \\\\\to V_{\ coffee} \times (T_2 - T_1)_{\ coffee} = - V_{\ milk} \times (T_2 - T_1)_{\ milk}.............(b)$

by replace the equation values (b) for calculating milk volumes as shown as follows:

$\to V_{\ coffee} \times (T_2 - T_1)_{\ coffee} = - V_{\ milk} \times (T_2 - T_1)_{\ milk}\\\\\to 250 ml \times (-10 .0^{\circ} \ \ C) = - V_{\ milk} \times (68.0^{\circ} \ \ C)\\\\\to V_{\ milk}= \frac{250 ml \times (-10 .0^{\circ} \ \ C) }{- (68.0^{\circ} \ \ C)}$

               $= 36.76 \ ml$