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.

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