How much solar energy (kJ) would have to be transferred to a 185.0 foot length of asphalt highway that is?

Question

43.0 feet wide and 24.3 centimeters deep in order to raise the temperature 1.50 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
http://answers.yahoo.com/question/index;_ylt=AqusAR8xr2oMMPnVaeu_CwTsy6IX?qid=20070124191938AATBMpA
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 = 185.0 ft. = 56.4 meters
Width = 43.0 ft. = 13.1 meters
Depth = 24.3 cm = .243 meters
Volume = (56.4 m long) * (13.1 m wide) * (.243 m deep)
Volume = 179.5 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) * (179.5 m^3)
Mass = 129420 kg


We now know, or are given, all the information we need to solve this problem.
Mass = 129420 kg
Specific heat = .920 kJ/kg °C
(Delta T) = 1.50 °C

Q = (129420 kg) * (.920 kJ per kg per °C) * (1.50 °C)
Q = 178600 kJ

So it would take about 1.79 E5 kJ of energy.

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

Answer 2

192.0 ft x 0.3048 m/ft = fifty 8.fifty two m 40-one.5 ft x 0.3048 m/ft = 12.sixty 5 m 21.a million cm / a hundred cm/m = 0.211 m fifty 8.fifty two x 12.sixty 5 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 a million.50 C = 155000 kJ

Answer 3

43 ft = 13.1064 meters
185 feet = 56.38800 meters

56.388 * 13.1064 * 0.243 = 179.587 m^3

721 * 179.587 = 129482.670 kg

0.920 * 129482.670 = 119124.057 kJ/oC

119124.057 * 1.50 = 178686.085 kJ


Just look at the units, and cancel out the ones you can.