Respuesta :
Answer:
[tex]q=120731598.109\ J=1.2\times 10^8\ J[/tex]
Explanation:
Given:
time of shower usage each day,[tex]t=9\ min[/tex]
flow rate through the shower, [tex]\dot V_s=1.8\ gal.min^{-1}[/tex]
volume of water used for washing each day, [tex]V_w=23\ gal[/tex]
initial temperature of water, [tex]T_i=60\ F=15.56\ ^{\circ}C[/tex]
final temperature of water, [tex]T_f=110\ f=43.33\ ^{\circ}C[/tex]
Now the total volume of water used each day:
[tex]V=\dot V_s\times t+V_w[/tex]
[tex]V=1.8\times 9+23[/tex]
[tex]V=39.2\ gal=148.3881\ L[/tex]
We know that the mass of 1 L of water is 1 kg, so the mass of water used each day:
[tex]m=148.3881\ kg[/tex]
Heat energy required for heating the water each day within the given temperature range:
[tex]Q=m.c.(T_f-T_i)[/tex]
where:
c = specific heat capacity of water = [tex]4184\ J.kg^{-1}.K^{-1}[/tex]
[tex]Q=148.3881\times 4184\times (43.33-15.56)[/tex]
[tex]Q=17247371.1584\ J[/tex]
Now the energy required per week:
[tex]q=Q\times 7[/tex]
since there are 7 days in a week
[tex]q=17247371.1584\times 7[/tex]
[tex]q=120731598.109\ J=1.2\times 10^8\ J[/tex]
Answer:
Explanation:
watr used for shower = 1.8 gallon per minute
Total water used for shower in 9 minutes , V = 1.8 x 9 = 16.2 gal per day
Water used for washing, V' = 23 gal per day
initial temperature of water, T1 = 60 F = 15.56 °C
final temperature of water, T2 = 110 F = 43.33 °C
Total Volume of water used per day = (23 + 16.2) x 3.785 litre
total mass, m = 148.372 kg
Heat required, H = mass x specific heat of water x rise in temperature
H = 148.372 x 4200 x (43.33 - 15.56)
H = 1.73 x 10^7 Joule
