The temperature in the freezer compartment of a household refrigerator is kept a

Question

The temperature in the freezer compartment of a household refrigerator is kept at 0 F. On a summer day the ambient temperature is 92 F. The effective performance factor for the refrigerator (qc/w where qc is the heat removed from the frozen food compartment and w is the electrical work required) is only one-tenth of that achievable by an ideal Carnot refrigerator. If electric power costs 10 cents per/kWh (1 watt is 1 joule for 1 second), what is the cost of making one hundred 50-g ice cubes? Assume that the heat of fusion for ice is 80 cal/g and that when the ice cube trays are put in the refrigerator, the water temperature is 77 o F. The specific heat capacity of ice is 0.5 cal/g K, half that of liquid water.

Explanation / Answer

Efficiency of the refrigerator is given by

n= 1- ( temp of freezer /ambient temp)

temp of freezer= 0 F= 255.32 K

ambient temp = 92 F= 306.48 K

so n= 1- (255.32/306.48)   

=(1 -0.833 )

=0.167

so efficiency is 16.7 %

given The effective performance factor for the refrigerator (qc/w) is only one-tenth of that achievable by an ideal Carnot refrigerator.

so q/w= 1/10 x 0.167

q/w = 0.0167

w=59.88q

heat in converting the water at 77 F into ice at 0 F for one 50 g ice cube is given by

q= mL + mSice dt1 +mSwdt2

whee L is the latent heat of fusion of ice

s specific heat capacity of ice

mass = 50 g

given water temp = 77 F= 298 K

so q= 50 x 80 + 50 x 1x ( 298-273) + 50 x 0.5 ( 273- 255.32)

=4000 + 1250 + 442

=5692 cal

heat for 100 cubes is

total q= 100 x 5692

=569200 cal

= 569200 x 4.2 J

q=2390640 J

w=59.88q

= 143152095.8

Given 10 cents per kwh

1 Kwh= 3600000 J

so 10 cents for 3600000 J

let b cents for 143152095.8 J

b= 397.64 cents

b= 3.97 dollars

cost is 3.97 dollars

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