In order to solve for this equation for heat capacity, do I need to treat it as a system of equations?

Question

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:

1. $q=m × 4.184 J/(g*C) × ΔT$
2. $q_{warm} + q_{cool} + q_{calorimeter} = 0$
3. $q_{calorimeter} = - (q_{warm} + q_{cool} ) = C_{calorimeter} × ΔT$.

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:

1. In the first equation, what does $C$ designate.
2. If it designates the heat capacity of the system, do I solve for $q$ by solving a system of equations?

Answer 1

In the first equation, what does C designate.

Since the final value is energy in joules, then it should be the the unit for temperature. She should have used °C to be clear.

If it designates the heat capacity of the system, do I solve for q by solving a system of equations?

No, because you don't have the correct equations. You need the amount of heat released from the hot water minus the amount of heat absorbed by the cold water to give you the heat absorbed by the calorimeter.