➢➣An ice plant produces 100 tons of ice in 24 hrs. If the plant uses water at
20

^{0 }C,**find its cooling capacity.**Specific heat and latent heat of solidification of water are 4.2 kJ/kg/K and 334 kJ/kg respectively.

**Solution:**

Given,

m = 100 tons = 100000 Kg

T = 20

^{0}C = 293 K
Specific heat, c

_{f}= 4.2 kJ/kg/K
Latent heat of solidification, h

_{fg}= 334 kJ/kg
Cooling Capacity =?

We know that,

Heat removed from water to ice at 20

^{0}C,**Q**

_{1}= m c_{f}T
= 100000 x 4.2 x 293

=1.23 x10

^{8}**KJ**
Total latent heat of freezing,

**Q**

_{2}= m h_{fg}
= 100000 x 334

= 3.34
x10

^{7}**KJ**
Total heat removed in 24 hours,

**Q = Q**

_{1}+ Q_{2 }
= 1.23 x10

^{8}+3.34 x10^{7 }^{ }= 156.46 x10

^{6}

**KJ**

So, Heat removed in 1 minutes

= 156.46 x10

^{6}/(24 x 60)
= 108.62 x10

^{3}**KJ/min**
Cooling Capacity,

= 108.62 x10

^{3}/210
=
517.39

**TR**

**So, Cooling Capacity of ice plant is = 517.39 TR**
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