Coffee Problem

Coffee Problem

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First watch the Introductory Video and Support Information on the upper left corner to learn how to use this simulation.

Then solve the following problem.

During the summer after your first year in college, 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 65.0°C. You then need to add enough milk at 4.00°C to drop the temperature of the coffee, initially at 95.0°C, to the ordered temperature.

Estimate the volume of milk and coffee:

Calculate the amount of milk (in ml) you must add to reach this temperature.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.


Heat transferred q, can be calculated as q = m c DT, where m is the mass of matter, c is the specific heat capacity in J/(g ºC), DT is the temperature change (DT = final temperature – initial temperature).


Heat released by coffee can be expressed as:

q coffee=mcoffeeccoffeeDTcoffee,

where DTcoffee= 65oC – 95oC = -30oC

mcoffee= Vcoffeedcoffee


Heat absorbed by milk can be expressed as:

q milk=mmilkcmilkDTmilk ,

where DTmilk= 65oC – 4oC = 61oC

mmilk= Vmilkdmilk


Since the coffee mug is insulating, the overall heat transferred within the mug is zero. That is,

q coffee= – q milk.


Now we have: mcoffeeccoffeeDTcoffee = – q milk =mmilkcmilkDTmilk

Plug DTcoffee= -30oC,  DTmilk= +61oC, mmilk= Vmilkdmilk, and mcoffee= Vcoffeedcoffee into the above equation we got: Vcoffeedcoffeeccoffee(-30oC)  = – Vmilkdmilkcmilk(61oC)


Since we assume dcoffee = dmilk = 1.000 g/mL and ccoffee = cmilk = 4.184 J/(g ºC), we can divide both sides of the above equation by -ccoffeedcoffee ( or  -dmilkcmilk) to simplify the equation as: Vcoffee(30oC)  = Vmilk(61oC)


Since the ordered coffee has a total volume of 200 mL, we can express Vcoffee  as:

Vcoffee = 200 – Vmilk


Plug Vcoffee = 200 – Vmilk into Vcoffee(30oC)  =Vmilk(61oC), we got


(200 – Vmilk) (30oC)  = Vmilk(61oC)


Solve the above equation for Vmilkand caluateVcoffee. Show your work on Data Sheet.

Experimental procedure:

  1. Place a bottle of hot coffee, a bottle of milk, and a 0.2L foam cup (in Glassware→Other) on the work bench. The foam cup is thermal insulating so it won’t lose any heat.
  2. Transfer the estimated amount of hot coffee and cold milk into the foam cup. Click on the foam cup to read the temperature of its content. Make sure you will read the temperature with two decimal places with in the information window (not the in the orange box). The information around the foam cup is not as accurate.















  1. According to the temperature of the coffee that you just made, make a new plan to make the temperature to be exactly 65.00 oC and volume to be exactly 200.00 mL, just as what the eccentric chemistry professor ordered precisely. (J).
  2. Once you obtained the precise coffee, screen shot the information window showing the temperature 65.00 oC. Turn in the screen shot with your lab report.
  3. And record the volume of hot coffee and the volume of cold milk that you used in Data Sheet. Use two decimal places for the volumes.



Data Sheet:

  1. Show your work to solve the following equation:

(200 – Vmilk) (30oC)  = Vmilk(61oC)




  1. What are the volume of hot coffee and the volume of cold milk that you used?


Post Lab Assignments:

  1. How much heat is required to heat a 18.4 g ice cube from –23.0 °C to –1.0 °C? Specific heat of ice is 108J/g °C. Show your work.











  1. How much would the temperature of 275 g of water increase if 36.5 kJ of heat were added? Specific heat of water is 4.184 J/g ° Show your work.










  1. A 45-g aluminum spoon (specific heat 0.88 J/g °C) at 24 °C is placed in 180 mL (180 g) of water at 85 °C and the temperature of the two become equal. Specific heat of water is 4.184 J/g ° What is the final temperature when the two become equal?Show your work.














  1. Upload the screen shot of the temperature and volume of the perfect coffee, completed Data Sheet, and answers to post lab questions to Canvas.

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