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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)
Solution
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 (Cw) = 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 = mw × Lmelt
= 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
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