liquid ammonia is used in ice factory for making ice from water .if water at 20degree celcious is to be converted into 2kg ice at 0 degree Celsius, how many grams of ammonia are to be evaporated
Answers
Mass of Ice = 2 kg.
Thus, Mass of water at 20°C = 2 kg.
Specific Heat Capacity of Water = 4200 J/kg°C
Change in temperature(Δt) = (20 - 0)
= 20 °C
Thus, Amount of Heat Energy Required to convert water at 20° to 0° C
= mass × Specific Heat Capacity × Change in temperature
= 20 × 4200 × 20
= 400 × 4200
= 168 × 10⁴ J.
Latent Heat of Fusion of Ice = 336000 J/kg.
Amount of Heat Energy Required to Convert water into ice at 0°
= mass × Latent Heat of Fusion
= 20 × 336000
= 672 × 10⁴ Joule
Thus, Total Heat Energy = 168 × 10⁴ + 672 × 10⁴
= 10⁴(168 + 672)
= 10⁴ × 800
= 8 × 10⁶ J.
Let the Mass of the Ammonia be m kg.
We know the Latent heat of Vaporization of Ammonia = 1400 × 10³ J/kg.
By the Principle of Calorimetry,
Heat Given By m kg of ammonia in Vaporization = Heat Taken by the Water to change into Ice.
⇒ m × 1400 × 10³ = 8 × 10⁶
⇒ m = (8/1400) × 10³
⇒ m = 5714.3 g.