Final Temperature of Ice-Water Mixture – Pre-Lab Physics Question

Pre-Lab Physics: Thermodynamics Question

Question:

Pre-Lab Question 53: A quantity of ice (250 grams at -5°C) is placed in 300 grams of water at 25°C. What is the final temperature of the mixture? What is the nature of the mixture?

(Hint: Calculating the final temperature is a little different in this scenario. It should be consistent with your answer to the latter question. Also note that ice and liquid water have different properties — both are provided in the table above.)

Answer:

Step 1: Use the energy balance principle

The energy required by the ice comes from the thermal energy released by the warm water. We’ll use:

Q = m × c × ΔT

Specific heat of ice: 2.1 J/g·°C
Specific heat of water: 4.18 J/g·°C
Latent heat of fusion (Lf): 334 J/g

Step 2: Heat required to warm the ice from -5°C to 0°C

Q₁ = m × c × ΔT = 250 g × 2.1 J/g·°C × 5°C = 2625 J

Step 3: Heat required to melt all the ice at 0°C

Q₂ = m × Lf = 250 g × 334 J/g = 83500 J

Total energy needed by ice

Q_{ice total} = Q₁ + Q₂ = 2625 J + 83500 J = 86125 J

Step 4: Energy released by water cooling from 25°C to 0°C

Q_water = m × c × ΔT = 300 g × 4.18 J/g·°C × 25°C = 31350 J

Step 5: Compare energy available and required

Since 31350 J (energy from water) is less than 86125 J (required to melt all the ice), not all ice will melt.

Therefore, the final temperature will be 0°C, and the mixture will consist of ice and water in equilibrium.

✅ Final Answer:

Final Temperature: 0°C
Nature of the Mixture: Ice and water in equilibrium

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