Physics two find the cost for Part A Problem 16.69 Part A Eigure 1) shows a hous
ID: 1405404 • Letter: P
Question
Physics two find the cost for Part AProblem 16.69 Part A Eigure 1) shows a house whose walls consist of plaster Find the total heating cost for the month. Express your answer using two significant figures 045), R-11 fberglass insulation, plywood (R 0.65), and oedar shingles (R-0.55). The roof has the same construction except it uses R-30 fiberglass insulation. The average outdoor temperature in winter is 23.F.and the houses mantained at 75.The houses oil furnace produces 100,000 Btu for every gallon of oil, and oil costs $2.20 per gallon. Assume that the house has 10 single-glazed windows, each measuring 2.8 ft by 4.7 ft. Four of the windows are on the south and admit sclar energy at the average rate of 30 Btu/h ft2 Al the windows lose heat; their R-factor is 0.90. S/mo My Answers Give Up Part B How much is the solar gain worth? Express your answer using two significant figures. Figure 1 of A-(36 ) h-14 ft × tan 30 My Answers Give Up Incorrect; Try Again; 5 attempts remainingg i0 ft Provide Feedback 6 ft
Explanation / Answer
Since you have asked to solve Part A, here's the solution -
The R-factors for the wall materails sum to give
Rwall = 0.45 + 11 + 0.65 + 0.55 = 12.65
R(roof) = 0.45 + 30 + 0.65 + 0.55 = 31.65
The perimeter of the house measures
2 * 28 ft + 2 * 36 ft = 128 ft
so, the 10-ft vertical walls have area 1280 ft^2.
There are also the triangular gables. Since there are two of them, each with area 1/2 b h, they give another bh or (28 ft)(14 ft * tan 30) = 226 ft^2.
So, A(wall) = 1506 ft^2
These R-12.65 walls lose 1/12.65 Btu/h/ft^2/F.
With 1506 ft^2 and a temperature difference of 52 degree F, the total heat loss rate through the walls is
H(wall) = (1/12.65 Btu/h/ft^2/F) (1506 ft^2)(52 degree F)
= 6190.671 Btu/h
The area of the pitched roof is larger than that of a flat roof by the factor 1/cos30, so the heat loss rate through the roof is
H(roof) = (1/31.65 Btu/h/ft^2/F) (36ft)(28ft)(52)/ cos30
= 1925.713 Btu/h
The total heat loss rate is then 8116.3846 Btu/h.
In a month, this results in a heat loss of Q = (8116.3846) (30 days/month) (24 h/day) = 5.84 MBtu.
Now for the oil, With 10^5 Btu (0.1 MBtu) per gallon, we will burn 58.4 gallons per month to produce that 5.84 MBtu. At $2.20/gal, that will cost $128.48.