The motor in a refrigerator has a power output of 400 w. if the freezing compartment is at 273 k and the outside air is at 306 k, assuming ideal efficiency, what is the maximum amount of heat (in joules) that can be extracted from the freezing compartment in 80.0 minutes?

15,883.63 KJ

Explanation:

Given

Power=400 W

Freezing compartment temperature is 273 K

Outside air Temperature=306 K

Time =80 minutes

Energy Required to deliver 400 w power in 80 minutes

E=1920 KJ

and we know COP of refrigerator is given by

Therefore 15,883.63 KJ is removed in 80 minutes

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz