How much solar energy (kJ) would have to be transferred to a 167.0 foot length of asphalt highway that is 44.5 feet wide and 22.1 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-oCEnter .
Do I calculate the volume of the region? I am unsure. Can someone guide me.|||Yeah. First calculate the Volume
Multiply the 3 dimension and get:
167ft(1m/3.281ft) x 44.5ft(1m/3.281ft) x 22.1cm(1m/100cm)
Volume = 152.66m^3
Now get the Mass
Mass = Volume x Density
Mass = 152.66m^3 x 721kg/m^3
Mass = 110,066.8 kg
Now you want to find out the amount of solar energy in kJ to raise the temp by 1C.
Q = mc(deltaT)
Q = 110,066.8kg(0.920kJ/kg-C)(1C)
Q = 101,261.45 kJ|||Yes, Volume is required to obtain the total mass of Asphalt.
167ft x 12in/ft x 2.54cm/in = 5,090cm = 50.9m
44.5ft x 12 x 2.54 = 1,356cm = 13.6m
22.1cm = 0.221m
0.221 x 13.6 x 50.9 = 153m鲁
153m鲁 x 721kg/m鲁 = 110,302kg of Asphalt.
S.H. = 0.920kJ/kg/掳C...Temp. increase = 1掳C
110,302kg x 0.920kJ/kg/掳C x 1掳C
= 101,478kJ (101.478MJ of Heat Energy (Solar).
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