Exams › NEET › Physics

When 1 kg of ice at 0°C melts to water at 0°C, the resulting change in its entropy, taking latent heat of ice to be 80 cal/g, is

1. 273 cal/K
2. 8 × 10⁴ cal/K
3. 80 cal/K
4. 293 cal/K

Correct answer: 273 cal/K

Solution

The change in entropy is given by ΔS = Q/T, where Q is the heat absorbed and T is the temperature in Kelvin. For 1 kg of ice, Q = 80 cal/g × 1000 g = 80000 cal. T = 273 K. Thus, ΔS = 80000/273 ≈ 293 cal/K.

Related NEET Physics questions

• An instrument used to measure high temperature is
• Ocean currents are caused by the setting up of in water
• For the same mass, which one of the following has the maximum thermal capacity?
• The excess temperature of a body falls from \( 12^{\circ} \mathrm{C} \) to \( 6^{\circ} \mathrm{C} \) in 5 minutes, then the time to fall the excess temperature from \( 6^{\circ} \mathrm{C} \) to \( 3^{\circ} \mathrm{C} \) is (assume the Newton's cooling is valid)
• Bunty mixed 440 gm of ice at \( 0^{\circ} \mathrm{C} \) with \( 540 \mathrm{gm} \) of water at \( 80^{\circ} \mathrm{C} \) in a bowl. Then what would remain after sometime in the bowl?
• Mercury boils at \( 356^{\circ} \mathrm{C} \). However mercury thermometers are made such that they can measure temperatures up to \( 500^{\circ} \mathrm{C} \). This is done by :

⚔️ Practice NEET Physics free + battle 1v1 →