Pre-Lab Calorimetry Problem – Final Equilibrium Temperature
Question:
A 140 gram piece of iron initially at 100°C is placed into 150 grams of water initially at 25°C.
What is the final equilibrium temperature (Tf) of the system, assuming no heat is lost to the environment?
Answer and Detailed Explanation:
Step 1: List the Given Values
- Mass of iron: mFe = 140 g
- Initial temperature of iron: TFe,initial = 100°C
- Specific heat of iron: cFe = 0.450 J/g°C
- Mass of water: mH2O = 150 g
- Initial temperature of water: TH2O,initial = 25°C
- Specific heat of water: cH2O = 4.18 J/g°C
Step 2: Use Principle of Heat Exchange
Heat lost by hot iron = Heat gained by cold water
mFe × cFe × (TFe,initial − Tf) = mH2O × cH2O × (Tf − TH2O,initial)
Step 3: Substitute Known Values
140 × 0.450 × (100 − Tf) = 150 × 4.18 × (Tf − 25)
63 × (100 − Tf) = 627 × (Tf − 25)
Step 4: Expand and Solve
Expand both sides:
6300 − 63Tf = 627Tf − 15675
Bring like terms together:
6300 + 15675 = 627Tf + 63Tf21975 = 690Tf
Step 5: Solve for Final Temperature
Tf = 21975 / 690 = 31.84°C
✅ Final Answer:
The final equilibrium temperature of the system is 31.84°C.
The final equilibrium temperature of the system is 31.84°C.