For a lab, we took the temperature of 15ml of tap water, $q_{cool}$, then added 20ml of tap water warmed to 80 degrees Celsius, $q_{warm}$.
We recorded a baseline temperature of 24.8, a temperature of 56 after adding the water, and a temperature of 42.9 100 seconds later.
I have the following equations:
- $q=m × 4.184 J/(g*C) × ΔT$
- $q_{warm} + q_{cool} + q_{calorimeter} = 0$
- $q_{calorimeter} = - (q_{warm} + q_{cool} ) = C_{calorimeter} × ΔT$.
In the first formula $ 4.184 J/(g*C)$ is the specific heat of water.
To the best of my knowledge, specific heat is usually expressed in grams-Kelvin; however, the TA may have provided the info in grams-Celsius (their units are equal, so I presume they're the same thing). However, she also uses $C$ to designate heat capacity as she does in the third equation.
If, in the first equation, $C$ designates heat capacity, then I have three equations, and three unknowns: $q_{calorimeter}, q_{warm}$, and $q_{cool}$, and I'd treat it as a system of three equations and solve for them.
However, if, in the first equation, $C$ designates 'Celsius' then I could just plug in the values for $q_{warm}$ and $q_{cool}$ and solve for $q_{calorimeter}$.
Two related questions:
- In the first equation, what does $C$ designate.
- If it designates the heat capacity of the system, do I solve for $q$ by solving a system of equations?