Answer to Question #108448 in General Chemistry for Galilea

Question

A 12.00 ounce cup of coffee from Starbucks is determined to have a temperature of 90.0 oC. The customer asks the barista to add just enough refrigerated milk (T = 4.00 oC) to make the cup of coffee have a final temperature of 70.0 oC. Calculate this volume of milk (d = 1.0 g/mL) required. Heat capacity of coffee = 4.18 J/g oC. Heat capacity of milk is 3.93 J/g oC.
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Answer 1

1) If 12 oz = milk + coffee

12 oz = 340.19 g

let the m (grams) be the mass of milk added, then mass of coffee is 340.19-m

Milk absorbs heat from coffee, so considering initial temperature as 4 ^oC and final temperature as 70 ⁰C (because milk and coffee are mixed together and mixture of coffee and milk has constant temperature),

the amount of heat absorbed by milk is m*(70-4)*3.93 = 259.38m (J)

The initial temperature of coffee is 90 ⁰C, final temperature is 70 ⁰C, then amount of heat given off by coffee is (340.19-m)*(90-70)*4.18 = 83.6*(340.19-m)=28439.884-83.6m (J)

The amount of heat given off by coffee and absorbed by milk is equal, then

259.38m = 28439.884 - 83.6m

28439.884 = 342.98m

m = 82.9 g $\approx$ 83 g

V = m/$\rho$=83/1=83 mL

2) If 12 oz = just coffee, with extra room for milk

12 oz =340.19 g

Let the m (grams) be the mass of milk added

Milk absorbs heat from coffee, so considering initial temperature as 4 ^oC and final temperature as 70 ⁰C (because milk and coffee are mixed together and mixture of coffee and milk has constant temperature),

the amount of heat absorbed by milk is m*(70-4)*3.93 = 259.38m (J)

The initial temperature of coffee is 90 ⁰C, final temperature is 70 ⁰C, then amount of heat given off by coffee is 340.19*(90-70)*4.18 = 28439.884 (J)

The amount of heat given off by coffee and absorbed by milk is equal, then

259.38m = 28439.884

m = 109.64 $\approx$ 110 (g)

V=m/$\rho$ = 110/1 = 110 mL