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: Latent heat of vaporisation of ammonia= 341 cal/g)

Answer

To convert 2 kg water at `20 ^@C` into ice at `O ^@C` completely, 586.5 g of ammonia should be evaporated

Explanation

Given: To find: Flow of heat: Mass of water/ice (m) `= 2 kg = 2 xx 10^3 g `
Change in temperature of water `(Delta T)=20-0=20^@C`
Latent heat of vaporisation of ammonia `L_("vap")_("ammonia") = 341 cal//g `
We know that Specific heat of water `C_w=1 cal//g^@C`, Latent heat of melting of ice `=L_("melt")= 80 cal//g`
To find : Mass of ammonia (M).
Flow of heat: Loss of heat :
`underset((20^@C))"water"overset(O_2)rarrunderset((0^@C))"water"overset(o_2)rarrunderset((0^@C))"ice"`
Gain of heat:
`underset("(liquid)")"Ammonia"overset(O_3)rarrunderset("(gas)")"Ammonia"`
Formulae: i. Heat energy to be given off in cooling water from `20^@C "to"0^@C(Q_1)=mxx C_W xx Delta T`
ii. Heat energy to be given off to convert water at `0^@C` into ice at `0^@C(Q_2)=m(L_("melt"))`
iii. Heat energy to be taken off using ammonia `(Q_3)=M(L_("vap"))_("ammonia")`
Calculation: From formula (i),
`Q_1=2 xx 10^3 xx 1xx (20-0)`
`=40 xx 10^3 cal `
From formula (ii).
`Q_2 =2 xx 10^3 xx 1 xx (20-0)`
`= 40xx 10^3 cal`
Total energy given off ,
`therefore Q_3 =Q_1 +Q_2`
`=(40 +160)xx 10^3`
`=200 xx 10^3 cal`
According to principle of heat exchange, this heat energy is to be taken away using 'M' mass of liquid ammonia. From formula (iii)
`therefore 200 xx 10^3=M xx 341`
`therefore M=(200 xx 10^3)/(341)=586.5 g `