Paragraph r Styles Problem 5: (10 pts) In a one hour, period, 2.00 kg of liquid

Question

Paragraph r Styles Problem 5: (10 pts) In a one hour, period, 2.00 kg of liquid water (initially at 0'C) is converted to ice (also at o'c) inside a freezer. The coefficient of performance for this freezer is 5.70 and it sits in a room at temperature 20°C. For water, L,-2.26 × 106 and Lt-3.33 x 105 kg kg a) How much (electrical) energy does the refrigerator require to operate for an hour? b) How much heat is exhausted into the room? c) If this is an ideal freezer, what is the temperature inside the freezer?

Explanation / Answer

a) Since Lf is the energy required for converting ice to water, Q1= 3.33*10^5*2 J is required in this process in 1 hour

Or, Q1= 666kJ

Given that coefficient of performance= 5.7, work required(which is in the form of electrical energy here)= 666/5.7= 116.84kJ

b) Heat exhausted into the room= 666kJ+ 116.84kJ= 782.84kJ

c) For ideal freezer, COP= Tin/ (Tout-Tin)

Or, 5.7= Tin/ 293-Tin

Or, Temperature inside the freezerTin= 249.268K= -23.880C

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