Question

7.3 Dulong and Petit- Report Form the Date: Name. Section/Group: Data Trial 1 Trial 2 Copper Mass of copper metal Mass of dry calorimeter 111 199 g|は4eal Mass of calorimeter + water Mass of water o C Water temperature-initial Copper temperature (hot)-initial o C Water Temperature -final Copper Temperature -final o c Unknown Trial 1 Trial 2 Mass of unknown sample

How to find the joules with the calorimeter? I only have the mass of the dry and the mass of the calorimeter with water.

Answer 1

Hi,
To know the heat released and / or absorbed is part of the law of conservation of energy which is established by the following equation:

Q = m * Ce * ΔT

Where m is the mass of the water, Ce the capacity to heat the water, this is a constant (4184 J / g ° C) and ΔT (Tc-Tf), it is the difference of the temperature of the hot water minus the final temperature in the calorimeter, thus obtaining the amount of heat (Q) that is transferred from the hot liquid, or the one that absorbs the cold (exothermic heat)

Trial 1
m = 39.47 g
Tc = 101.1 ° C
Tf = 25.5 ° C

Q = 39.47 g * 4184 J / g ° C (101.1 - 25.5) ° C = 12,484,771.5 J

Trial 2
m = 38.61 g
Tc = 101.1 ° C
Tf = 26.2 ° C

Q = 38.61 g * 4184 J / g ° C (101.1 - 26.2) ° C = 12,099,663.6 J

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