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 stops in every day and orders 250 ml of house coffee at precisely 95*C. He then adds enough milk at 4*C to drop the temperature of the coffee to 90*C.
Calculate the amount of milk, in ml, the professor must add to reach this temperature. Show all of your work including equations and units. Assume coffee and milk have the same specific heat capacity of 4.186 J/g*C and density of 1.0 g/ml.

Step 1: Write down what is given to you in the problem
Initial temp of coffee = 95*C
Final temp = 90*C
Specific heat capacity of coffee and milk= 4.186 J/g*C
Density = 1.0 g/ml
Initial Temp of milk = 4*C

Step 2: Calculate how much energy the coffee loses when being cooled down from 95*C to 90*C. (q = C x m x Δ T) (3 points)





Step 3: Calculate the mass of milk needed to cool down the coffee using the same equation as in step 2 (you must rearrange it to solve for mass). The change in temperature is going to be the final temp of the coffee/milk mixture minus the temp of the milk. (4 points)






Step 4: Use the density of the milk to calculate to milliliters of milk needed. This is your answer. (1 point)

Respuesta :

Answer:

did you ever get the answer to this?

Explanation:

The amount of milk that the Professor must add to reach the temperature is 18.09 ml

Step 1 : List the given data

The amount of coffee ordered = 250 mL

Initial temperature of coffee = 95°C

Temperature of milk = 4°C

Final temperature of coffee = 90°C

Specific heat capacity = 4.186 J/g°C

density = 1.0 g/ml

Assumptions :

x = amount of milk added

Initial amount of coffee = ( 250 - x )

Step 2 : Amount of energy released/gained

heat released by coffee ( q1 ) = m*c * ΔT

= 1.0(250 - x ) * 4.86 * ( 90 - 95 ) ------ ( 1 )

heat gained by milk ( q2 ) = m* c * ΔT

where;  m = ( density * volume of milk ) , C= 4.186 J/g°C , ΔT = ( 90 - 4 ) = 86°C

q2 = 1.0( x ) * 4.186 * 86 ------- ( 2 )

Step 3: Determine mass of milk needed in milliliters

equate equations (1)  and (2) and solve

1.0(250 - x ) * 4.86 * ( 90 - 95 ) = 1.0( x ) * 4.186 * 86

(250 - x ) * 4.86 * - 5 = x * 360

= ( 250 - x ) * -24.3 = 360 x

= 6075 + 24.3 x = 360 x

∴ x ( amount of milk added ) = 6075 / 335.7 = 18.09 mL

Hence the amount of milk added by the professor = 18.09 mL

to learn more : https://brainly.com/question/15682825

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