How much thermal energy is required to heat all of the water in a swimming pool by 1°C if the dimensions are 4 ft deep
by 20 ft wide by 75 ft long? Report your result in megajoules.
1ft=0.3048m1ft=0.3048m1ft=0.3048m
volume of water in the pool =(75×0.3048)(20×0.3048)(4×0.3048)=169.90m3=(75\times 0.3048)(20\times 0.3048)(4\times 0.3048)=169.90m^3=(75×0.3048)(20×0.3048)(4×0.3048)=169.90m3
Density of water =1000kg/m3=1000kg/m^3=1000kg/m3
Mass of water =169.90×1000=169.90\times 1000=169.90×1000
Heat energy =mcΔT=(169.90×1000)×4200×1=713.58×106J=713.58MJ=mc \Delta T = (169.90\times 1000)\times 4200\times 1=713.58 \times 10^6J=713.58MJ=mcΔT=(169.90×1000)×4200×1=713.58×106J=713.58MJ
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