i don't know what am i doing worng.. plz help?

Question

How much solar energy (kJ) would have to be transferred to a 178.0 foot length of asphalt highway that is 44.5 feet wide and 21.5 centimeters deep in order to raise the temperature 3.50 oC ?


The average density of asphalt is 721 kg/m3


The specific heat of asphalt is 0.920 kJ/kg-oC

Answer 1

Compute the mass of the asphalt
1 in =2.54 cm

1ft=.3048 m

Volume
44.5*.3048*
178*.3048*
.215 m3
=158.2 m3

using density, calculate mass
=158.2*721
=114073.2101
energy is
m*specific heat*temp rise
=3.5*0.920*112073

=367316 kJ

j

Answer 2

First, you figure the volume of asphalt, that is length by width by thickness. We have to convert the length and width in m to be consistant, so that we have
178/3.2808399 * 44.5/3.2808399 *0.215 = 158.215 m^3
With a density of 721 kg/m^3, this means we have 114073 kg of asphalt.
Since we have to increase the temperature by 3.5 C, and the specific heat is 0.92 kJ/kg-C, this means 3.22 kJ for each kg of asphalt, so in the end, the amount of energy needed is 367315.7 kJ.

Answer 3

Q = m x c x ΔT
m = ro x V, V = abc = 178.0 x 0.3048 m x 44.5 x 0.3048 m x 0.215 m = 158.215 m3
m = 721 kg/m3 x 158.215 m3 = 114073 kg
Q = 114073 kg x 0.920 kJ/kgK x 3.50 K = 3.67 x 10^5 kJ