Answer to Question #108634 in General Chemistry for Alex

Question

A12.00 ounce cup of coffee from Starbucks is determined to have a temperatureof 90.0oC. The customer asksthe barista to add just enough refrigerated milk(T = 4.00oC)to make the cup of coffee have a final temperature of 70.0oC.Calculate thisvolume of milk (d = 1.0 g/mL) required.Heat capacityof coffee = 4.18 J/g oC. Heat capacity of milk is 3.93 J/g oC.

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 ⁰C 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≈ 83 g

V = m/ρ

ρ=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 ⁰C 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 ≈ 110 (g)

V=m/ρ

ρ = 110/1 = 110 mL