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

RONALD E B192.0 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

DandelionRONALD 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