Joe Smith, the chemical engineer from Acme Chemical Company, wanted to determine the final volume in a large glass cylinder is he dropped a 12.52 pound ingot of lead whose density was 112.3 g/cm^3 into the cylinder which already contained 2500. mL of water. What is the final volume of water in the cylinder.
volume of lead is
12.52lb * 453.59g/lb * 1cm^3/112.3g = ? cm^3
Add that to the 2500 cm^3 already there.
Technically, the final volume of water is still 2500mL, but I think they wanted to know the total volume of contents.
Check earlier problem. diameter not radius was given
To find the final volume of water in the cylinder, we need to consider the change in volume caused by adding the lead ingot.
First, let's convert the weight of the lead ingot from pounds to grams. We know that 1 pound is equal to 453.6 grams. Therefore, the weight of the ingot in grams would be:
12.52 pounds * 453.6 grams/pound = 5670.67 grams
Next, we can find the change in volume caused by adding the lead ingot by using its density. The density of lead is given as 112.3 grams per cubic centimeter (g/cm^3). Therefore, the volume (V) of the lead ingot can be calculated by dividing its mass (m) by its density (d):
V = m / d = 5670.67 g / 112.3 g/cm^3 = 50.53 cm^3
Since the lead ingot is dropped into a cylinder containing 2500 mL of water, we need to convert this volume of water from milliliters to cubic centimeters because the volume of the lead ingot has been calculated in cubic centimeters. We know that 1 mL is equal to 1 cm^3.
Therefore, the volume of water in cubic centimeters is also 2500 cm^3.
To find the final volume of water in the cylinder, we need to add the initial volume of water to the change in volume caused by adding the lead ingot:
Final volume of water = Initial volume of water + Change in volume
Final volume of water = 2500 cm^3 + 50.53 cm^3 = 2550.53 cm^3
Therefore, the final volume of water in the cylinder is 2550.53 cm^3.