Question

use a graduated cylinder to add approximately mL of water to the calorimeter. Measure the mass of the calorimeter (no lid) and water (iron) to the nearest 0.01. (the mass of the cylinder is 12.42, figured that out already.

Answer 1

I can help — I just need one clarification: when you say “the mass of the cylinder is 12.42,” is that the empty (tare) mass of the graduated cylinder, or the mass of the cylinder when filled with water? Also, what volume (mL) of water did you actually add to the calorimeter?

While you get that, here’s the standard procedure and the formulas you’ll use (all masses to be reported to 0.01 g):

Procedure
1. Weigh the empty calorimeter (no lid). Record M_cal_empty (g).
2. Fill the graduated cylinder with the desired volume of water (read volume in mL). If you prefer, you can transfer water directly to the calorimeter.
3. If using the cylinder-mass method: weigh the cylinder when filled and record M_cyl_full (g). You already have the empty cylinder mass M_cyl_empty (12.42 g) if that is indeed the empty mass.
4. Transfer the water from the cylinder to the calorimeter.
5. Weigh the calorimeter now containing water (and no iron yet) if possible and record M_cal_water (g).
6. Place the iron (or other solid) into the calorimeter and weigh the system. Record M_cal_water_iron (g).
7. If you need the iron mass, weigh the iron separately on its own (preferable) and record M_iron (g). If you didn’t weigh it separately, you can get it by difference (see formulas).

Key formulas
- Mass of water (from cylinder): mass_water = M_cyl_full − M_cyl_empty
- Mass of water (from calorimeter difference): mass_water = M_cal_water − M_cal_empty
- Mass of iron (if not weighed separately): mass_iron = M_cal_water_iron − M_cal_water

Notes
- 1 mL of (room-temperature) water ≈ 1.00 g (use mass from weighing when possible — that’s more accurate than converting volume).
- Report every mass to the nearest 0.01 g.
- If you tell me which numbers you have (M_cal_empty, M_cyl_full, M_cal_water, M_cal_water_iron, or the volume added), I’ll do the arithmetic and give the masses rounded to 0.01 g.

Example (illustration)
- M_cyl_empty = 12.42 g
- M_cyl_full = 62.42 g → mass_water = 62.42 − 12.42 = 50.00 g (so you added 50.00 mL)
- M_cal_empty = 86.15 g
- After adding water: M_cal_water = 136.15 g → mass_water = 136.15 − 86.15 = 50.00 g (consistency check)
- After adding iron: M_cal_water_iron = 140.34 g → mass_iron = 140.34 − 136.15 = 4.19 g

Send me the masses/volume you measured and I’ll compute and round them for you.