how much heat is required to warm 1500 gallons of water from 70 to 140 degrees F?

heat= mass*sepcificheat capcity*changetemp

I don't know if you can find the specific heat capacity for water in SAE units. So here it is :

specific heat capacity water= 1BTU./lb-degF

now convert 1500 gallons water to lbs...That depends on temperature, so take the starting temperature 70F, compute mass, it wont change

http://www.engineeringtoolbox.com/water-density-specific-weight-d_595.html

To calculate the amount of heat required to warm 1500 gallons of water from 70 to 140 degrees Fahrenheit, you would need to use the heat capacity formula. The formula is:

Q = m * C * ΔT

Where:
Q = Heat required (in calories or joules, depending on the unit of heat capacity)
m = Mass of water (in grams or kilograms, depending on the unit of heat capacity)
C = Specific heat capacity of water (in calories per gram per degree Celsius or joules per kilogram per degree Celsius)
ΔT = Change in temperature (in degrees Celsius or Kelvin)

First, let's convert the gallons of water to liters, as it is a more common unit in the heat capacity formula.

1 gallon = 3.78541 liters

So, 1500 gallons of water is equal to:
1500 * 3.78541 = 5678.115 liters

Now, let's convert the temperature difference from Fahrenheit to Celsius:

ΔT = 140°F - 70°F = 70°F

To convert Fahrenheit to Celsius, you can use the formula:

°C = (°F - 32) / 1.8

So, the temperature difference in Celsius is:
ΔT = (140°F - 32) / 1.8 - (70°F - 32) / 1.8 ≈ (60) / 1.8 = 33.33°C

Next, we need to find the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 joules per gram per degree Celsius, or 1 calorie per gram per degree Celsius.

Now, since we have the temperature difference in Celsius, we can calculate the amount of heat required using the formula:

Q = m * C * ΔT

Let's assume we will use the specific heat capacity of water in joules:

Q = 5678.115 kg * 4.18 J/g°C * 33.33°C

Calculating this would give you the amount of heat required in joules.