How much solar energy (kJ) would have to be transferred to a 186.0 foot?

Question

length of asphalt highway that is 46.0 feet wide and 24.6 centimeters deep in order to raise the temperature 2.50 oC ?



The average density of asphalt is 721 kg/m3



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

Answer 1

Your question is virtually identical to this question,
http://answers.yahoo.com/question/index;_ylt=Ao7WYuxmV5BHCYM5PblLzA_sy6IX?qid=20070124100533AAXBWND
And the process to solve it is the same.


The heat energy needed to be absorbed by the asphalt (Q) is equal to the product of the mass of the asphalt (m), the specific heat of the asphalt (c), and the change in temperature it experiences (delta T),

Q = m * c * (delta T)


We are given the dimensions of a rectangular solid piece of asphalt from which we can calculate its volume.
Volume = length * width * depth
Length = 186.0 ft. = 56.7 meters
Width = 46.0 ft. = 14.0 meters
Depth = 24.6 cm = .246 meters
Volume = (56.7 m long) * (14.0 m wide) * (.246 m deep)
Volume = 195.3 m^3 of asphalt.

We are given the density of the asphalt.
We should know that,
Density = mass / volume,
So we can find the mass of the asphalt.

Mass = Density * Volume
Mass = (721 kg/m^3) * (195.3 m^3)
Mass = 140811 kg


We now know, or are given, all the information we need to solve this problem.
Mass = 140811 kg
Specific heat = .920 kJ/kg °C
(Delta T) = 2.50 °C

Q = (140739 kg) * (.920 kJ per kg per °C) * (2.50 °C)
Q = 323865 kJ

So it would take about 3.24 E5 kJ of energy.

http://www.digitaldutch.com/unitconverter/

Answer 2

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