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?

Answer 1

`Raise = 1391.85\mu F`

Step-by-step explanation:

Given

`Temperature = 20^(\circ) F`

`Gas = 30ft^3`

`Efficiency = 90\%`

Required

Determine the pounds of water

`1031BTU = 1ft^3` of water

`1\mu F = 20BTU` - The energy to raise 1 pound of water by 20F

The raise in pounds is then calculated as:

`Raise = Efficiency * Energy * Volume(gas) * cubic\ of\ water`

This gives:

`Raise = 90\%* (1\mu F)/(20BTU) * 30ft^3 * (1031BTU)/(1ft^3)`

`Raise = 90\%* (1\mu F)/(20BTU) * 30 * (1031BTU)/(1)`

`Raise = 90\%* (1\mu F)/(20) * 30 * (1031)/(1)`

`Raise = 90\%* (1\mu F)/(20) * 30 * 1031`

`Raise = 90\%* (1\mu F)/(20) * 30930`

`Raise = 90\%* 1\mu F * 1546.5`

`Raise = 1\mu F * 1391.85`

`Raise = 1391.85\mu F`