3. How many pounds of water could be raised in temperature 20°F by the 90% efficient burning of 30 cubic feet of natural gas?

1 answer
Question:

3. How many pounds of water could be raised in temperature 20°F by the 90% efficient burning of 30
cubic feet of natural gas?

Answers

Answer:[tex]Raise = 1391.85\mu F[/tex]Explanation:Given[tex]Temperature = 20^{\circ} F[/tex][tex]Gas = 30ft^3[/tex][tex]Efficiency = 90\%[/tex]RequiredDetermine the pounds of water[tex]1031BTU = 1ft^3[/tex] of water[tex]1\mu F = 20BTU[/tex] - The energy to raise 1 pound of water by 20F The raise in pounds is then calculated as:[tex]Raise = Efficiency * Energy * Volume(gas) * cubic\ of\ water[/tex]This gives:[tex]Raise = 90\%* \frac{1\mu F}{20BTU} * 30ft^3 * \frac{1031BTU}{1ft^3}[/tex][tex]Raise = 90\%* \frac{1\mu F}{20BTU} * 30 * \frac{1031BTU}{1}[/tex][tex]Raise = 90\%* \frac{1\mu F}{20} * 30 * \frac{1031}{1}[/tex][tex]Raise = 90\%* \frac{1\mu F}{20} * 30 * 1031[/tex][tex]Raise = 90\%* \frac{1\mu F}{20} * 30930[/tex][tex]Raise = 90\%* 1\mu F * 1546.5[/tex][tex]Raise = 1\mu F * 1391.85[/tex][tex]Raise = 1391.85\mu F[/tex]

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