Final Temperature of Water and Ice Mixture – Rankers Physics

Calorimetry: Practice Problem & Solution

300 gm of water at \(25^\circ\text{C}\) is added to 100 g of ice at \(0^\circ\text{C}\). The final temperature of the mixture is:
\(-\frac{5}{3}^\circ\text{C}\)
\(-\frac{5}{2}^\circ\text{C}\)
\(5^\circ\text{C}\)
\(0^\circ\text{C}\)

Solution Explained:

To solve this problem, we apply the core principles of Calorimetry. Understanding the underlying formula is key to arriving at the correct answer below:

Heat lost by water to reach \(0^\circ\text{C}\) is \(300 \times 1 \times 25 = 7500\text{ cal}\). Heat required to melt all ice is \(100 \times 80 = 8000\text{ cal}\). Since heat lost is less than heat required to melt all ice, some ice remains, and the final temperature is \(0^\circ\text{C}\).

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