Calculating Volume and Mass of Iron Block Using Equal-Arm Balance | Scales

[tandoorichicken]

A block of iron is suspended from one end of an equal-arm balance by a thin wire. To balance the scales, 2.35 kg are needed on the scale pan at the other end.
(a) What is the vloume of the block?
(b) Next a beaker of water is placed so that the iron block is submerged in the beaker but not touching the bottom. What mass is now necessary to balance the scales?

What I did:
density of iron = 7.87 * 10^3 kg/m^3
a) 2.35/$\rho_{\text{iron}}$ = V = 299 cm^3
b) V = V of displaced water = m of water in grams
2.35 kg - 299 g = 2.05 kg

 

[HallsofIvy]

Looks good to me.

 

[M98Ranger]

of water

Your calculations for the volume and mass are correct. However, it is important to note that the density of iron may vary depending on its purity and temperature. Additionally, it is more common to use units of kilograms or grams for mass and cubic meters or centimeters for volume in scientific measurements. Therefore, the volume of the iron block can also be expressed as 0.000299 m^3 or 299 cm^3, and the mass of the water needed to balance the scales can be written as 2.05 kg or 2050 g.