# If 10 g of ice at 0°C is converted to water at the same temperature, the change in entropy will be- (latent heat 80 cal/g)

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If 10 g of ice at 0°C is converted to water at the same temperature, the change in entropy will be- (latent heat 80 cal/g)
1. 3.29 cal/K
2. 2.93 cal/K
3. 29.3 cal/K
4. 32.9 cal/K

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Correct Answer - Option 2 : 2.93 cal/K

Concept:

Entropy

• It is a measure of disturbance, the higher the entropy with the positive sign, the more are the chances of the reaction taking place spontaneously.

$S = \frac{{{\rm{Δ }}Q}}{T}$

Where S is entropy, Δ Q is heat exchanged and T is the temperature in Kelvin.

Latent Heat

• The heat required to change the state of matter without changing the temperature is called Latent Heat.
• Latent heat of fusion that is heat required by 1 gram of water to raise its temperature by 1 Kelvin or 1 °C is called latent heat of fusion which is

l = 80 calorie.

Calculation:

Given mass of ice = 10 gm

Heat required to convert in water = m × l = 10 gm × 80 cal = 800 Cal

Temperature T = 0°C = 273 K

$S = \frac{{{\rm{Δ }}Q}}{T}$

$\implies S = \frac{800\ cal}{273K}$

⇒ S = 2.93 cal/K

So, the correct option is 2.93 cal/K