Question

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

Answer 1

Answer:

$Raise = 1391.85\mu F$

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:

$Raise = 90\%* \frac{1\mu F}{20BTU} * 30ft^3 * \frac{1031BTU}{1ft^3}$

$Raise = 90\%* \frac{1\mu F}{20BTU} * 30 * \frac{1031BTU}{1}$

$Raise = 90\%* \frac{1\mu F}{20} * 30 * \frac{1031}{1}$

$Raise = 90\%* \frac{1\mu F}{20} * 30 * 1031$

$Raise = 90\%* \frac{1\mu F}{20} * 30930$

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

$Raise = 1\mu F * 1391.85$

$Raise = 1391.85\mu F$