How many tons of refrigeration are necessary to cool 10 imperial gallons of water from 80°F to 40°F in 60 minutes?

To determine how many tons of refrigeration are necessary to cool 10 imperial gallons of water from 80°F to 40°F in 60 minutes, we can use the formula for calculating the cooling capacity required. This involves taking into account the volume of water, the change in temperature, and the specific heat capacity of water.

First, we note that 1 imperial gallon of water weighs approximately 10 pounds. Therefore, 10 imperial gallons would weigh around 100 pounds.

Next, we calculate the amount of heat that needs to be removed to cool the water. The specific heat of water is about 1 British thermal unit (BTU) per pound per degree Fahrenheit. The temperature change (ΔT) from 80°F to 40°F is 40°F.

Using the formula for calculating the total BTUs required to cool the water:

Total BTUs = Weight of water (lbs) × Specific heat (BTU/lb°F) × Temperature change (°F)

Total BTUs = 100 lbs × 1 BTU/lb°F × 40°F = 4000 BTUs.

Next, to find out how many tons of refrigeration this amount of BTUs represents, we need to know how many BTUs one ton