The outer walls of your house have an R-value of 70.0 K m 2 / W, and a total are

Question

The outer walls of your house have an R-value of 70.0 K m2 / W, and a total area of 1900 m2. Lets assume that, from the beginning of December to the end of February, the temperature outside the house is 0°C while you set your thermostat to maintain a constant temperature of 22°C inside the house.

(a) How much energy is conducted through the walls of your house in this three-month period? Assume it is not a leap year.
__J

(b) How much energy would be conducted through the walls if you lowered the thermostat so as to maintain a constant temperature of 20°C?
__J

(c) Every kW-h (kilowatt-hour) of energy costs 20 cents. First, convert 1 kW-h to joules, and then determine how much money you would save by keeping your thermostat at 20°C for the three months.
__dollars

Please show all work!

Explanation / Answer

let,

area of heat passing A=1900m^2

R=l/k=70 k.m^2/w

temperature differnce dT=22 oC

time t=90 days

a)

P=(k*A*dT/l)

E/t=(A*dT)/(l/k)

E/t=1900*(22)/(70)

E=597.143*(90*24*60*60)

E=4.64*10^9 J

b)

if dT=20 oC

E/t=(A*dT)/(l/k)

E/t=1900*(20)/(70)

E=542.857*(90*24*60*60)

E=4.22*10^9 J

c)

1kwh=3.6*10^6 J

energy cost =20 cents for 1kwh

if

E1=4.64*10^9 J

E1=4.64*10^9/(3.6*10^6)

E1=1288.9kwh

now,

cost of E1 is =1288.9

cost of E1 is =25777.78 cents

and

if

E2=4.22*10^9 J

E2=4.22*10^9/(3.6*10^6)

E2=1172.22kwh

now,

cost of E2 is =1172.22*20 cents

cost of E1 is =23444.44 cents

difference in cost is=(25777.78-23444.44)

=2333.34 cents

or

=23.33 dollars

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