Question

10 kg of ice at \(-10^\circ\text{C}\) is added to 100 kg of water to lower its temperature from \(25^\circ\text{C}\). Consider no heat exchange to surroundings. The decrement in the temperature of water is ________ \( ^\circ\text{C} \). (Specific heat of ice \(=2100\,\text{J kg}^{-1}\!^\circ\text{C}^{-1}\), specific heat of water \(=4200\,\text{J kg}^{-1}\!^\circ\text{C}^{-1}\), latent heat of fusion of ice \(=3.36\times10^5\,\text{J kg}^{-1}\))

• A. \(15\)
• B. \(6.7\)
• C. \(11.6\)
• D. \(10\)

Hint:

Always account for latent heat before equating temperature changes in ice–water mixing problems.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This is a calorimetry problem. Heat lost by water equals heat gained by ice to reach a final equilibrium temperature \(T\).
Step 2: Detailed Explanation:
Heat gained by ice to reach final temperature \(T\):
- To warm to \(0\ ^\circ\text{C}\): \(Q_1 = m_i c_i \Delta T = 10 \cdot 2100 \cdot 10 = 2.1 \times 10^5\ \text{J}\).
- To melt at \(0\ ^\circ\text{C}\): \(Q_2 = m_i L = 10 \cdot 3.36 \times 10^5 = 33.6 \times 10^5\ \text{J}\).
- To warm melted water to \(T\): \(Q_3 = m_i c_w T = 10 \cdot 4200 \cdot T = 42000T\).
Total \(Q_{gain} = 35.7 \times 10^5 + 42000T\).
Heat lost by original water:
\(Q_{loss} = m_w c_w (25 - T) = 100 \cdot 4200 \cdot (25 - T) = 4.2 \times 10^5(25 - T) = 105 \times 10^5 - 420000T\).
Equating \(Q_{gain} = Q_{loss}\):
\(35.7 \times 10^5 + 42000T = 105 \times 10^5 - 420000T\)
\(462000T = 69.3 \times 10^5 \Rightarrow T = 15\ ^\circ\text{C}\).
Decrement = \(25 - 15 = 10\ ^\circ\text{C}\).
Step 3: Final Answer:
The decrement in temperature is \(10\ ^\circ\text{C}\).