Question

A camper is about to drink his morning coffee. He pours 400 grams of coffee, initially at into a 250-g aluminum cup, initially at 16.0°C. What is the equilibrium temperature of the coffee-cup system, assuming no heat is lost to the surroundings? The specific heat of aluminum is 900 J/kg°C. Assume that the specific heat of coffee is the same as the specific heat of water.

Answer 1

For the coffee, we are given that:
m = 0.4 kg
Ti = 75 degrees celcius
C = 4186 J/Kg.K

For the aluminum container, we are given that:
m = 0.25 kg
Ti = 16 degrees celcius
C = 900 J/Kg.K

Now, the rule that we will use to solve this problem is:
∑Q = 0
∑Q = Qcoffee + Qcontainer 
where: Q = CmΔT
Therefore:
∑Q = (4186)(0.4)(Tfinal - 75) + (900)(0.25)(Tfinal - 16) = 0
1674 (Tfinal - 75) + 225 (Tfinal - 16) = 0
1674 Tfinal - 125550 + 225 Tfinal - 3600 = 0
1899 Tfinal = 129150
Tfinal = 68.009 dgrees celcius