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Question: Supercooling is a state where liquids do not solidiby even below their normal freezing point. Superc...

Supercooling is a state where liquids do not solidiby even below their normal freezing point. Supercooled water can remain in the liquid state as low as -40°C. If a small piece of ice is dropped then water suddenly freezes. If super cooled water at -10°C suddenly starts freezing then in a very short interval a mixture of water and ice both at 0°C is formed. If 1 kg of water freezes then 334 kJ of energy is released. Specific heat of ice = 2.1 kJ kg−1^{-1}k−1^{-1}, specific heat of water (in supercooled state also) = 4.2 kJ kg−1^{-1}K−1^{-1}, Latent heat of ice at 0°C = 334 kJ kg−1^{-1}. Consider 1 kg of water and mark the correct statements :-

A

Latent heat of fusion at -10°C is nearly 313 kJ kg−1^{-1}.

B

Latent heat of fusion at -10°C is 334 kJkg−1^{-1}.

C

If supercooled water suddenly freezes at -10°C then approximately 0.67 kg of ice is formed.

D

If supercooled water suddenly freezes at -10°C then approximately 0.13 kg of ice is formed.

Answer

A, D

Explanation

Solution

The latent heat of fusion (LfL_f) can be approximated as a function of temperature using the specific heats of the liquid (cliquidc_{liquid}) and solid (csolidc_{solid}) phases: Lf(T)=Lf(T0)+∫T0T(cliquid−csolid)dTL_f(T) = L_f(T_0) + \int_{T_0}^{T} (c_{liquid} - c_{solid}) dT

Statement (A): Latent heat of fusion at -10°C is nearly 313 kJ kg−1^{-1}. Given Lf(0∘C)=334L_f(0^\circ C) = 334 kJ/kg, cwater=4.2c_{water} = 4.2 kJ kg−1^{-1}K−1^{-1}, and cice=2.1c_{ice} = 2.1 kJ kg−1^{-1}K−1^{-1}. Let T0=0∘CT_0 = 0^\circ C and T=−10∘CT = -10^\circ C. Lf(−10∘C)=Lf(0∘C)+(cwater−cice)×(T−T0)L_f(-10^\circ C) = L_f(0^\circ C) + (c_{water} - c_{ice}) \times (T - T_0) Lf(−10∘C)=334 kJ/kg+(4.2−2.1) kJ kg−1K−1×(−10−0) KL_f(-10^\circ C) = 334 \text{ kJ/kg} + (4.2 - 2.1) \text{ kJ kg}^{-1}\text{K}^{-1} \times (-10 - 0) \text{ K} Lf(−10∘C)=334+(2.1)×(−10)=334−21=313 kJ kg−1L_f(-10^\circ C) = 334 + (2.1) \times (-10) = 334 - 21 = 313 \text{ kJ kg}^{-1}. Thus, statement (A) is correct.

Statement (B): Latent heat of fusion at -10°C is 334 kJkg−1^{-1}. As calculated above, the latent heat of fusion at -10°C is approximately 313 kJ kg−1^{-1}, not 334 kJ kg−1^{-1} (which is the value at 0°C). Thus, statement (B) is incorrect.

Statements (C) and (D): If supercooled water suddenly freezes at -10°C then approximately 0.67 kg of ice is formed or approximately 0.13 kg of ice is formed. When supercooled water at -10°C suddenly freezes, it releases energy. This energy causes the temperature of the remaining water to rise to 0°C, and then the phase change (freezing) occurs at 0°C, resulting in a mixture of ice and water at 0°C. Let micem_{ice} be the mass of ice formed. The remaining mass of water is (1−mice)(1 - m_{ice}) kg. The energy released by freezing micem_{ice} kg of water at 0°C is Qreleased=mice×Lf=mice×334Q_{released} = m_{ice} \times L_f = m_{ice} \times 334 kJ. This energy is absorbed by the (1−mice)(1 - m_{ice}) kg of water to heat it from -10°C to 0°C: Qabsorbed=(1−mice)×cwater×ΔT=(1−mice)×4.2 kJ kg−1K−1×(0−(−10)) KQ_{absorbed} = (1 - m_{ice}) \times c_{water} \times \Delta T = (1 - m_{ice}) \times 4.2 \text{ kJ kg}^{-1}\text{K}^{-1} \times (0 - (-10)) \text{ K} Qabsorbed=(1−mice)×4.2×10=(1−mice)×42Q_{absorbed} = (1 - m_{ice}) \times 4.2 \times 10 = (1 - m_{ice}) \times 42 kJ. Equating the energy released and absorbed: mice×334=(1−mice)×42m_{ice} \times 334 = (1 - m_{ice}) \times 42 334mice=42−42mice334 m_{ice} = 42 - 42 m_{ice} 376mice=42376 m_{ice} = 42 mice=42376≈0.1117m_{ice} = \frac{42}{376} \approx 0.1117 kg. This value is approximately 0.13 kg. Thus, statement (D) is correct. Statement (C) is incorrect as 0.67 kg is not close to 0.1117 kg.