Question

A graduated cylinder contains 155 ml of water. a 15.0-g piece of iron (density = 7.86 g/ml) and a 20.0-g piece of lead (density = 11.3 g/ml) are added. what is the new water level, in milliliters, in the cylinder?

Answer 1

The new water level in the cylinder is 158.676 mL.

Step-by-step explanation:

The initial volume of water in the graduated cylinder is 155 mL. When the 15.0-g piece of iron is added, it will displace an amount of water equal to its own volume. To calculate the volume of the iron, we divide its mass by the density of iron:

Volume of iron = Mass of iron / Density of iron

Volume of iron = 15.0 g / 7.86 g/mL = 1.906 mL

The same calculation is done for the 20.0-g piece of lead:

Volume of lead = Mass of lead / Density of lead

Volume of lead = 20.0 g / 11.3 g/mL = 1.770 mL

The total volume of water after adding the iron and lead is the initial volume of water plus the volumes of the iron and lead:

New water level = Initial water level + Volume of iron + Volume of lead

New water level = 155 mL + 1.906 mL + 1.770 mL = 158.676 mL

The new water level in the graduated cylinder is 158.676 mL.

Answer 2

Initial volume of water is 155 mL. Convert the mass of iron and lead into volume using density values as follows:

d=\frac{m}{V}

Here, d is density, m is mass and V is volume of objects

volume of piece of iron can be calculated as:

V=\frac{m}{d}=\frac{15 g}{7.86 g/mL}=1.90 mL

Similarly, volume of lead can be calculated as:

V=\frac{m}{d}=\frac{20 g}{11.3g/mL}=1.77 mL

To calculate the final volume of water, volume of both iron and lead should be added to the initial volume of water.

V_{final}=V_{initial}+V_{iron}+V_{lead}=155 mL+1.90 mL+1.77mL=158.67 mL

Thus, new water level in milliliters will be 158.67 mL.