Question

An 8.6 grams piece of aluminum, heated to 100°C, is placed in a coffee cup (styrofoam) calorimeter that contains 402.4 grams of water at 25°C. The specific heat of water is 4.18 J/g°C and the specific heat of aluminum is 0.900 J/g°C.

What will be the final equilibrium temperature for both the aluminum and the water? (Assume no heat is lost to the surroundings and the coffee cup absorbs a negligible amount of heat).

Answer 1

The final equilibrium T° will be 24.65°C

Step-by-step explanation:

A typical excersise of calorimetry, to apply this formula:

Q = m . C . ΔT

As no heat is lost to the surroundings, the heat that loses the aluminum and gains the water will be the same.

mass Al . C Al . ΔT = mass water . CH2O . ΔT

ΔT = T°final - T°initial

8.6 g . 0.900 J/g°C . (T°final - 100°C) = 402.4 g . 4.18J/g°c . (T°final - 25°C)

7.74 J/°C T°final - 774J = 1682.03 T°final - 42050.8J

-774 J + 42050.8J = 1682.03 T°final - 7.74 J/°C T°final

41276.8 J = 1674.29 J/°C T°final

41276.8 J /1674.29 °C/J =T°final

24.65°C = T°final