Question

Liquid ammonia is used in ice factory for making ice from water. If water at 20 °C is to be converted into 2 kg ice at 0 °C, how many grams of ammonia are to be evaporated? (Given: The latent heat of vaporization of ammonia= 341 cal/g) Answer : 586.4 g. Solve the problem

Answer 1

Let the mass of the ammonia evaporated by 'm' kg. 

Now, As per as the Question, Liquid ammonia is used to convert the water at 20° C into the 2 kg ice at 0° C. 

We know that the mass will never change during the sate of the matter. Therefore, mass of the water(m₁) is 2 kg. 
Specific Heat Capacity(c) = 4200 J/kg°C. 
Change in temperature(Δt) = (20 - 0) 
 = 20° C. 

∴ Energy released to decrease the temperature of the water from 20° C to the water at 0° C = m₁cΔt
   = 2 × 4200 × 20
   = 4200 × 40
   = 168000 J.

We know, Latent heat of the fusion of the Ice (L)  = 336000 J/kg.. 
Now, Energy released to convert the Water into Ice at 0° C = m₁L
= 2 × 336000
= 672000 J. 

Now, Total energy released while converting the water from 20° into ice at 0° is 168000 + 672000 = 840000 J.

Latent heat of the Vaporization of the Ammonia = 341 cal./g
= 1426744 J/kg. 
Now, Energy released by the 'm' kg of the ammonia = m × 1426744
 = 1426744m J. 


According to the Principle of the Calorimetry, 

Heat energy given by the Ammonia = Energy taken by the water to convert into ice. 

  ∴ 1426744m = 840000
⇒ m = 0.58875 kg. 
  $\boxed{m = 588.75 g.}$

[Note ⇒ The latent heat of the Vaporization of the Ammonia is given in the calories. On conversion, I have taken unit factor as 4.18. It may be possible that on conversion, the value may be 4.176 or something else. Thus, there may be the difference in the final answer. But such difference does not creates any problems.]


Hope it helps.

Answer 2

Hi friend
Here is your answer
Hope it helps you


plz Mark me down as a brainliest