Question

The outer walls of your house have an R-value of 76.0 K m $^2$ / W, and a total area of 2100 m $^2$ . Let’s 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.

How much energy is conducted through the walls of your house in this three-month period? Assume it is not a leap year. Note that a number like 1.23 x10^6 can be entered as 1.23e6
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Part (b)

The outer walls of your house have an R-value of 76.0 K m $^2$ / W, and a total area of 2100 m $^2$ . Let’s 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.

How much energy would be conducted through the walls if you lowered the thermostat so as to maintain a constant temperature of 20°C? Assume it is not a leap year. Note that a number like 1.23 x10^6 can be entered as 1.23e6
_______ J

The outer walls of your house have an R-value of 76.0 K m $^2$ / W, and a total area of 2100 m $^2$ . Let’s 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.

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.
$_______

Answer 1

Use the following equation to find the amount of heat transferred through the walls. QAT-I) TR R 2-4(1,-1) (2100 mº )(22-01( 90 days (74) (3905)) 76.0 Km/W = 4.727x10' 2.-47,-1,1% (2100 m)(20-01( 90 days ( 2 (1500 :)) 76.0 Km/W = 4.30x10° Convert the units of energy transferred from I to kWh as follows: Q.-(4.727x109 ) 1 -14727x109/1W-s/1kW-s 1h) 10W-s) 3600 s) =1.313x10 kWh Q: =(4.30x10°]) 1W-81kW-SY_1h) . =(430x10 15 10 W-s) 3600 s) =1.1937x10 kWh The total cost for heat transfer of Q is, cost(Q)=(1313x109 kW.h 20 cents (1 kWh cost(Q)=2.626x109 cents The total cost for heat transfer of Q.is, cost(Q)=(1.1937x10 kW.h 20 cents *" 1 kW.h) cost(Q)=2.3874x10* cents The amount of money can be saved by maintaining the inside temperature at 20 degrees Celsius instead of 22 degrees Celsius is, saved amount = (2.626x10* cents)-2.3874x10 cents) 0.01 dollar = 0.2386x10* cents 1 cent) = 23.865 After rounding off the value to 3 significant figures 23.9$