A 5.0-m-diameter garden pond is 0.400 m deep. Solar energy is incident on the pond at an average rate of 400W/m^2 . If the water absorbs all the solar energy and does not exchange energy with its surroundings, how many hours will it take to warm from 16.0 degrees C to 22.0 degrees C?
2007-11-18 01:54:01 · 2 answers · asked by gbutterfly339 1 in Science & Mathematics ➔ Physics
Power in is a product of radiation energy density and area
Pin= Rp A=
=400 pi D^2/4
Heat required to heat mass m of water (heat capacity Cp) from T1 to T2 is
Q=mCp(T2-T1) where mass m is a product of density and volume
m=pA d
p- density (for water 1E+3 kg/m^3)
A - area
d- depth of pond
Cp(water)= 4.18 E+3 Joules / (kg deg K)
Now total energy required is
Q=pA d Cp(T2-T1)
By definition power is a rate of energy flow
P=E/t=Q/t
t= Q/P
t=[pA d Cp(T2-T1)]/[Rp A]
t=[p d Cp(T2-T1)]/[Rp ]
t=[1e+3 x 0.4 x 4.18E+3 (6)/[400 ]
t=25080 s or 25,000 s (using proper SF)
t=6.97 =7.0 hours
2007-11-18 02:08:35 · answer #1 · answered by Edward 7 · 1⤊ 0⤋
Pin= 400 pi D^2/4
Heat e is
Q=mCp(T2-T1)
m=pA d
Q=pA d Cp(T2-T1)
P=E/t=Q/t
t= Q/P
2007-11-18 10:10:45 · answer #2 · answered by mj 3 · 0⤊ 0⤋