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

To find the decrement in the temperature of water when ice at \(-10^\circ\text{C}\) is added, we need to calculate the amount of heat absorbed by the ice as it warms to \(0^\circ\text{C}\), melts, and then further warms to the final equilibrium temperature. The total heat lost by the water should match this total heat absorbed by the ice due to the principle of conservation of energy (assuming no heat exchange with the surroundings).

Step-by-step calculation:

1. Heat required to raise the temperature of ice from \(-10^\circ\text{C}\) to \(0^\circ\text{C}\):

The specific heat of ice \(c_{\text{ice}}\) is \(2100\,\text{J kg}^{-1}\!^\circ\text{C}^{-1}\).

Q_1 = m_{\text{ice}} \cdot c_{\text{ice}} \cdot \Delta T_{\text{ice}} = 10 \cdot 2100 \cdot (0 - (-10)) = 21000 \,\text{J}

2. Heat required to melt the ice at \(0^\circ\text{C}\):

The latent heat of fusion \(L\) is \(3.36 \times 10^5 \,\text{J kg}^{-1}\).

Q_2 = m_{\text{ice}} \cdot L = 10 \cdot 3.36 \times 10^5 = 3.36 \times 10^6 \,\text{J}

3. Heat required to raise the temperature of melted ice (now water) to the final equilibrium temperature \(T\):

Let the final temperature decrement be \(\Delta T\).

Q_3 = m_{\text{melted water}} \cdot c_{\text{water}} \cdot \Delta T = 10 \cdot 4200 \cdot \Delta T \,\text{J}

Total heat absorbed by the ice:

Q_{\text{total, ice}} = Q_1 + Q_2 + Q_3 = 21000 + 3.36 \times 10^6 + 10 \cdot 4200 \cdot \Delta T \,\text{J}

4. Heat lost by the water lowering its temperature by \(\Delta T\):

The original water mass is 100 kg, and its specific heat is \(4200\,\text{J kg}^{-1}\!^\circ\text{C}^{-1}\).

Q_{\text{water}} = 100 \cdot 4200 \cdot \Delta T \,\text{J}

Setting the heat lost by water equal to the total heat gained by the ice:

100 \cdot 4200 \cdot \Delta T = 21000 + 3.36 \times 10^6 + 10 \cdot 4200 \cdot \Delta T

Simplifying the equation,

100 \cdot 4200 \cdot \Delta T - 10 \cdot 4200 \cdot \Delta T = 21000 + 3.36 \times 10^6

Simplify and solve for \(\Delta T\):

378000 \cdot \Delta T = 3.381 \times 10^6 \Delta T = \frac{3.381 \times 10^6}{378000} \approx 10

The decrement in the temperature of water is 10 \( ^\circ\text{C} \).