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 1

Given: 1) Change in temperature of water (ΔT) = 20° C 

2) mass of ice (m) = 2 kg = 2000 g/2 × 10³ g

3) `("L"_"vap")_"ammonia"` = 341 cal/g

4) Specific heat of wate (C_w) = 1 cal/g° C

5) `("L"_"melt")_"ice"` = 80 cal/g

Find: Mass of ammonia (M)

1) Amount of heat released in converting water at 20° C to water at 0°C (Q₁)

Q₁ = m × c × ΔT

= 2 × 10³ × 1 × 20

Q₁ = 40 × 10³ cal

2) Amount of heat released in converting water at 0° C into ice at 0° C (Q₂)

Q₂ = m_w × L_melt

= 2 × 10³ × 80

= 160 × 10³ cal

Total heat given out (Q₃) = Q₁ + Q₂

Q₃ = (40 + 160)10³ 

Q₃ = 200 × 10³ cal

According to principle of heat exchange - heat released by water is used by liquid ammonia

Q₃ = `"M""L"_(("vap")_("ammonia"))`

200 × 10³ = M × 341

M = `(200 xx 10^3)/341`

M = 586.5 g