Hi, I have a doubt. I'm doing my lab report: Thermochemistry and Specific Heat Capacity. I got stuck in the heat capacity. to solve this I have to use this formula qwater = mCp∆T, but I am not that specific heat capacity of water, I have to use 1calorie / gram ° C, or, 4.186 joules/gram ° C. Could you explain to me when I have to use each one? Here is my report, I don't need you to help me solve it, just explain my doubt please.
Thermochemistry
Thermochemistry can be considered as a branch of thermodynamics that deals with the connections between warmth, work, and various types of energy, formed because of different synthetic and actual cycles. Thermochemistry describes the energy changes that occur as a result of reactions or chemical changes in a substance.
Exergonic Reaction
The term exergonic is derived from the Greek word in which ‘ergon’ means work and exergonic means ‘work outside’. Exergonic reactions releases work energy. Exergonic reactions are different from exothermic reactions, the one that releases only heat energy during the course of the reaction. So, exothermic reaction is one type of exergonic reaction. Exergonic reaction releases work energy in different forms like heat, light or sound. For example, a glow stick releases light making that an exergonic reaction and not an exothermic reaction since no heat is released. Even endothermic reactions at very high temperature are exergonic.
Hi, I have a doubt. I'm doing my lab report: Thermochemistry and Specific Heat Capacity. I got stuck in the heat capacity. to solve this I have to use this formula qwater = mCp∆T, but I am not that specific heat capacity
of water, I have to use 1calorie / gram ° C, or, 4.186 joules/gram ° C.
Could you explain to me when I have to use each one? Here is my report, I don't need you to help me solve it, just explain my doubt please.
Part 1: Determination of the Heat Capacity of a Calorimeter
Water A
- Mass = 50.0 g
- Initial Temperature = 24.1°C
- Final Temperature = 28.7°C
- Specific heat capacity = 1 Cal /g °C
- Temperature change = 4.6 °C
- Heat of Water A (q) = 230 cal
Water B
- Mass = 50.0 g
- Initial Temperature: 36.7 °C
- Final Temperature = 28.7 °C
- Specific heat capacity = 1 cal /g °C
- Temperature change = -8.0 °C
- Heat of Water B (q) =
Calorimeter
- Heat of the Calorimeter =
- Initial Temperature = 24.1°C
- Final Temperature = 28.7 °C
- Temperature change = 4.6 °C
- Heat capacity (Kc) = __________________
Part 2: Determination of the Specific Heat Capacity of a Penny
Water
- Mass =40 g
- Initial Temperature =23.6 °C
- Final Temperature =24.6 °C
- Specific heat capacity = __________________
- Temperature change = ___________________
- Heat of water (q) = ____________________
Calorimeter
- Initial Temperature =23.6 °C
- Final Temperature = 6 °C
- Temperature change = ___________________
- Heat capacity (Kc) = __________________
- Heat of Calorimeter (q) = ________________
Penny
- Heat of Penny = ________________
- Mass =37.4 g
- Initial Temperature =45.6
- Final Temperature =______________
- Temperature change = ___________________
- Specific Heat Capacity of a Penny (Cp)= _________________
Thanks a lot
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