A camper pours 500 g of coffee, initially in a pot at 80 oC, into a 110-g aluminum cup initially at 30 oC. What is the final temperature? Assume that coffee has the same specific heat as water and that no heat is exchanged with the surroundings. (cwater=4.18 J/goC). (cAluminum=0.900J/g. C)

The final temperature : t=77.74 °C

Given

500 g coffee, t = 80 °C

110 g aluminum cup, t = 30 °C

Required

The final temperature

Solution

Q in = Q out

Q absorbed = Q released

Q aluminum cup = Q coffee

110 x 0.9 x (t-30)=500 x 4.18 x (80-t)

99t-2970=167200-2090t

2189t=170170

t=77.74 °C

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