Liquid ammonia is used in ice factory for making ice from water. If wa

1.	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 is to be evaporated?(Given: The latent heat of vaporization of 1 ammonia = 341 cal/g)
Answer» Data : m₁ = 2kg, ΔT₁ = 20 °C – 0 °C = 20 °C, c₁ = 1 kcal/kg·°C, L₁ (ice) = 80 kcal/kg, L2 (vaporization of ammonia) = 341 cal/g = 341 kcal/kg, m₂ =? Q₁ (heat lost by water) = m₁c₁ ΔT₁ + m₁L₁ = 2kg × 1 kcal/kg·°C × 20 °C + 2 kg × 80 kcal/kg = 40 kcal + 160 kcal = 200 kcal Q₂ (heat absorbed by ammonia) = m₂L₂ = m₂ × 34l kcal/kg According to the principle of heat exchange, Q₁ = Q₂ ∴ 200 kcal = m₂ × 341 kcal/kg ∴ m₂ = \(\frac{200}{341}\) kg = 0.5864 kg = 586.4 g 586.4 g of ammonia are to be evaporated.

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 is to be evaporated?(Given: The latent heat of vaporization of 1 ammonia = 341 cal/g)

Answer»

Data : m₁ = 2kg, ΔT₁ = 20 °C – 0 °C

= 20 °C, c₁ = 1 kcal/kg·°C, L₁ (ice) = 80 kcal/kg,

L2 (vaporization of ammonia) = 341 cal/g = 341

kcal/kg, m₂ =?

Q₁ (heat lost by water) = m₁c₁ ΔT₁ + m₁L₁

= 2kg × 1 kcal/kg·°C × 20 °C + 2 kg × 80 kcal/kg

= 40 kcal + 160 kcal = 200 kcal

Q₂ (heat absorbed by ammonia) = m₂L₂

= m₂ × 34l kcal/kg

According to the principle of heat exchange, Q₁ = Q₂

∴ 200 kcal = m₂ × 341 kcal/kg

∴ m₂ = \(\frac{200}{341}\) kg = 0.5864 kg = 586.4 g

586.4 g of ammonia are to be evaporated.

Discussion

No Comment Found

Related InterviewSolutions

Why are heaters fitted near the floor and air conditioners near the ceiling of a room?
Why are the Upper layers of water in a pond or a lake hotter than the lower layers during the hot summer?
In the arrangements A and B shown in Figure 4.7, pins P and Q are fixed to a metal loop and an iron rod with the help of wax. In which case are both the pins likely to fall at different times? Explain.
The heat energy required to raise the temperature of 20 kg of water from 25° C to 75° C. A) 103 cal B) 104 cal C) 105 cal D) 106 cal
Collect specific heats of various substances.
Calculate the two specific heats of nitrogen from the following data : `gamma = c_(p)//c_(v) = 1.51` density of nitrogen at N.T.P `= 1.234 g litre ^(-1)` and `J = 4.2 xx 10^(7) erg cal^(-1)`.
Calculate difference in specific heats for `1` gram of air at `N.T.P`. Given density of air at N.T.P. is `1.293 g litre^(-1), j=4.2xx10^(7)" erg "cal^(-1)`.
A vessel contains ice and is in thermal equilibrium at `-10^(@)C` and is supplied heat energy at the rate of 20 cal `s^(-1)` for 450 seconds. If the mass of ice is 0.1 kg and due to supply of heat energy, the whole ice just melts find the water equivalent of the vessel. (Take specific heat of ice `=0.5 cal g^(-1)""^(@)C^(-1)` and specific heat of the vessel is `=0.1 cal g^(-1)""^(@)C^(-1)`. Latent heat of fussion = `80 cal g^(-1)` and assume that no heat is transferred to the surroundings)
if specific heat capacity of mercury is 0.033 cal `g^(-1)" ".^(@)C^(-1)`, how much heat is gained by 0.05 kg of mercury when its temperature rises from `68^(@)F` to 313K?
Calculate the specific heat capacity at constant volume for a gas. Given specific heat capacity at constant pressure is `6.85 cal mol^(-1) K^(-1), R = 8.31 J mol^(-1)K^(-1)`. `J = 4.18 J cal^(-1)`.