At winter design conditions, a house is projected to lose heat at a rate of 60,000 Btu/h. The internal heat gain from people, lights, and appliances is estimated to be 6000 Btu/h. If this house is to be heated by electric resistance heaters, determine the requried rated power of these heaters in KW to maintain the house at constant temperature.

15.83 KW

Explanation:

Losing heat at a rate = 60000btu/h

Gaining heat at a rate = 6,000btu/h

The difference in heat levels

= 60000btu/h - 6000btu/h

= 54000btu/h

So we need to convert 54000btu/h to KW

1kw = 3412.142btu

? = 54000btu

We cross multiply

54000btu x 1kw = ?3412.142btu

So we divide through by 3412.142 to get the unknown

54000/3412.142

= 15.83kw

So we can conclude that 15.83kw is the required power of this heater to keep the house at a constant temperature

  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