What is the final Temperature of the mixture?

Question

25g cube of Al is heated to 150c and then added to 100g of water at 25c.
( sp. Heat of Al= 0.22 cal/g C )

Answer 1

Equilibrium means Q1 + Q2 = 0.

Therefore,

mass 1 * specific heat 1 * (Tf - Ti1) + mass 2 * specific heat 2 * (Tf - Tf1) = 0

Use equation:

m1 c1 (Tf - Ti1) + m2 c2 (Tf - Ti2) = 0

Given:

m1 = 25 g = 0.025 kg
m2 = 100 g = 0.1 kg
Ti1 = 150 C = 423.15 K
Ti2 = 90 C = 298.15 K
c of aluminum = 0.22 cal/(gC) = 920.48 J/(kgK)
c of water = 4186 J/(kgK)

Substitute:

(0.025)(920.48)(Tf - 423.15) + (0.1)(4186)(Tf - 298.15) = 0

Solve for Tf...

Tf = 304.664 K = 31.51 C

Note: I converted to SI units only because I'm most comfortable working in them. If all of your units match up, there's no need to convert.

Answer 2

Choose a reference temperature (again, 0°C is good since no phase transitions are involved) Calculate the energy needed to get the temperatures given:
25*1*100 = 2500 Cal for the water
150*.22*25 = 825 Cal for the aluminum.
That's a total of 3325 Cal. At equilibrium the water and aluminum will have the same temperature, so
T*1*100 + T*.22*25 = 3325 or
T = 3325/(1*100 + .22*25) = 33.25 °C above the reference temperature (which we chose to be zero)

You might try running the same calculation using, say, 10°C as the reference. Then the thermal energy in the water would be 1500 Cal. If you do it that way, you'll get a temperature of 23.25°C above the reference for a temperature of 33.25°C just as before.


Doug