How much does an 18 foot by 12 foot by 1 inch block of ice weigh?
The volume of an 18 foot by 12 foot by 1 inch block of ice can be calculated by multiplying its length, width, and height together and converting to cubic feet:

18 ft * 12 ft * (1/12) ft = 18 * 1 * 12 / 12 = 18 cubic feet

The weight of ice depends on its density, which can vary depending on temperature and other factors. However, a common approximation for the density of ice is 57.2 pounds per cubic foot.

Using this approximation, we can calculate the weight of an 18 foot by 12 foot by 1 inch block of ice as follows:

Weight = Volume * Density = 18 cubic feet * 57.2 pounds per cubic foot = 1,030.4 pounds

Therefore, an 18 foot by 12 foot by 1 inch block of ice weighs approximately 1,030.4 pounds.
Can you give me a formula to calculate the weight in ounces based on cubic inches of ice?