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: Identify the three stages of heat absorption by ice.
The ice at $-10^\circ$C must (i) warm up to $0^\circ$C, (ii) melt, and (iii) warm up as water to the final temperature $T_f$. Step 2: Heat gained by ice warming from $-10^\circ$C to $0^\circ$C.
\[ Q_1 = m_i \cdot s_{\text{ice}} \cdot \Delta T = 10 \times 2100 \times 10 = 2.1 \times 10^5 \text{ J} \] Step 3: Heat gained by ice melting at $0^\circ$C.
\[ Q_2 = m_i \cdot L = 10 \times 3.36 \times 10^5 = 3.36 \times 10^6 \text{ J} \] Step 4: Heat gained by melted ice warming from $0^\circ$C to $T_f$.
\[ Q_3 = m_i \cdot s_{\text{water}} \cdot T_f = 10 \times 4200 \times T_f = 42000\,T_f \text{ J} \] Step 5: Total heat absorbed by ice.
\[ Q_{\text{absorbed}} = Q_1 + Q_2 + Q_3 = 2.1\times10^5 + 3.36\times10^6 + 42000\,T_f = 3.57\times10^6 + 42000\,T_f \] Step 6: Heat lost by water cooling from $25^\circ$C to $T_f$.
\[ Q_{\text{lost}} = m_w \cdot s_{\text{water}} \cdot (25 - T_f) = 100 \times 4200 \times (25 - T_f) = 420000(25 - T_f) \] Step 7: Apply conservation of energy (no heat exchange with surroundings).
\[ Q_{\text{absorbed}} = Q_{\text{lost}} \] \[ 3.57\times10^6 + 42000\,T_f = 420000(25 - T_f) \] \[ 3570000 + 42000\,T_f = 10500000 - 420000\,T_f \] \[ 462000\,T_f = 6930000 \] \[ T_f = \frac{6930000}{462000} = 15^\circ\text{C} \] Step 8: Find the decrement in temperature of water.
\[ \Delta T_{\text{water}} = 25 - T_f = 25 - 15 = \boxed{10^\circ\text{C}} \]