Asked by tj
Check all items that you used, either directly or indirectly, each time you did this computation.
the first law of thermodynamics (heat lost by a metal equals the heat gained by the water)
the ability of heat to flow from a hot object to a cooler one
masses for the metal and the water in the calorimeter
temperature changes for the water and the metal
the known specific heat of the water
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All five — check each one. Briefly:
- the first law of thermodynamics (heat lost by a metal equals the heat gained by the water) — used to set q_metal + q_water = 0 (energy conservation).
- the ability of heat to flow from a hot object to a cooler one — used to assign signs and to know which body loses heat and which gains it.
- masses for the metal and the water in the calorimeter — used in q = m c ΔT for each substance.
- temperature changes for the water and the metal — ΔT values enter q = m c ΔT.
- the known specific heat of the water — the standard c_water (≈4.184 J/g·°C) is needed to compute q_water.
(If the calorimeter itself has a non‑negligible heat capacity you would also need its mass or heat capacity, but that was not listed.)
- the first law of thermodynamics (heat lost by a metal equals the heat gained by the water) — used to set q_metal + q_water = 0 (energy conservation).
- the ability of heat to flow from a hot object to a cooler one — used to assign signs and to know which body loses heat and which gains it.
- masses for the metal and the water in the calorimeter — used in q = m c ΔT for each substance.
- temperature changes for the water and the metal — ΔT values enter q = m c ΔT.
- the known specific heat of the water — the standard c_water (≈4.184 J/g·°C) is needed to compute q_water.
(If the calorimeter itself has a non‑negligible heat capacity you would also need its mass or heat capacity, but that was not listed.)
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