How much solar energy (kJ) would have to be transferred?
Question:
The average density of asphalt is 721 kg/m3
The specific heat of asphalt is 0.920 kJ/kg-oC
Answer:
The mass of asphalt is 44.5 ft * (.3048 m/ft) * 158 ft * (.3048 m/ft) * 18.3 cm * (0.01 m/cm) * 721 kg/m3 = 86815 kg
So the solar energy needed is h = m * c * dT = 86815 kg * .920 kJ/kg C * 4 C = 317,162 kJ
Someone just asked the same question. Not sure if it was you, but I answered it before but had forgot to convert the units so I re-edited it. But I'll copy/paste what I wrote for the same question here:
Q = m * c * delta T
m = density * volume
volume = l * w * h
197 ft. = 60 m
41.5 ft. = 12.65 m
So...
volume = 60 * 12.65 * 0.236 = 179.124 m^3
mass = 721 kg/m^3 * 179.124 m^3
= 129,148.4 kg
Q = (129,148.4 kg)(0.920 kJ/kgC)(4 C)
= 475,266.112 kJ
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