How much solar energy (kJ) would have to be transferred to a 192.0 foot length of asphalt highway that is 41.5 feet wide and 21.1 centimeters deep in order to raise the temperature 1.00 oC ?The average density of asphalt is 721 kg/m3The specific heat of asphalt is 0.920 kJ/kg-oC
192.0 ft x 0.3048 m/ft = 58.52 m
41.5 ft x 0.3048 m/ft = 12.65 m
21.1 cm / 100 cm/m = 0.211 m
58.52 x 12.65 x 0.211 = 156.2 m^3
156.2 m^3 x 721 kg/m^3 = 112619 kg
112619 kg x 0.920 kJ/kg C x 1.50 C = 155000 kJ
RONALD E B is correct, but he made one small error…
(192.0 ft x 0.3048 m/ft = 58.52 m
41.5 ft x 0.3048 m/ft = 12.65 m
21.1 cm / 100 cm/m = 0.211 m
58.52 x 12.65 x 0.211 = 156.2 m^3
156.2 m^3 x 721 kg/m^3 = 112619 kg)
112619 kg x 0.920 kJ/kg C x 1.00 C = 103609.48 kJ