How to Calculate Temperature Change in a Partially Exposed Water Calorimeter

[ccw]

Homework Statement

I've stumped myself on what (I think) should be a very easy problem. I'm trying to come up with an expression for the temperature of a small amount of water as heat is added at a constant rate. The water is partially exposed to open air, so can lose energy to the surroundings. The total temperature change is 20 degrees or so over many minutes, so I do need to keep track of the loss.

Homework Equations

The power is coming from a resistor, so the instantaneous energy change should be (I^2)R-k(T_water-T_air), where k accounts for exposed surface area, etc, etc and which I can determine as the water cools back down. This should be directly proportional to the instantaneous change in temperature, right?

I've baffled myself thoroughly enough that I'm currently just trying to determine two things. First, If I've described the instantaneous power in and out, should I be able to solve the differential equation easily by way of integration? Second, is it reasonable to try to extract the specific heat of water from a setup like this? That is, is there any way to fit a heating/cooling curve from a setup like this having only the specific heat as a free parameter without having very detailed information about surface area, airflow, etc?

If that was coherent enough to make sense of, I'd be extremely grateful for a kick in the right direction

 

[jbsteele]

.The Attempt at a SolutionWe can set up the following differential equation to determine the temperature of the water as heat is added at a constant rate: dT/dt = (I^2R - k(T_water-T_air))/mcwhere m is the mass of the water and c is its specific heat capacity.Integrating this equation yields: T_water = (I^2Rt)/mc + (k/mc)(T_air) + Cwhere C is an arbitrary constant of integration.It is possible to extract the specific heat of water from this equation by measuring the temperature of the water and the air, the resistance of the resistor, and the amount of current passed through it.

 

[Moetasim]

.I can provide the following response to the problem of calculating temperature change in a partially exposed water calorimeter:

Firstly, it is important to understand the basic principles of calorimetry. Calorimetry is the science of measuring the amount of heat transferred during a chemical or physical process. In a closed system, where no heat is lost to the surroundings, the change in temperature can be directly related to the amount of heat gained or lost. However, in an open system, like the one described in the problem, heat can be lost to the surroundings and this needs to be taken into account.

To calculate the temperature change in a partially exposed water calorimeter, we need to consider the heat gained by the water from the resistor and the heat lost to the surroundings. This can be expressed mathematically as:

Q_water = m_water * c_water * ΔT_water

where Q_water is the heat gained by the water, m_water is the mass of water, c_water is the specific heat of water, and ΔT_water is the change in temperature of the water.

Now, considering the heat lost to the surroundings, we can use Newton's law of cooling, which states that the rate of heat loss is directly proportional to the temperature difference between the object and the surroundings. This can be expressed mathematically as:

Q_loss = k * A * (T_water - T_air)

where Q_loss is the heat lost to the surroundings, k is a constant that takes into account the exposed surface area, A is the surface area of the water exposed to the surroundings, T_water is the temperature of the water, and T_air is the temperature of the surroundings.

Combining these two equations, we can write an expression for the change in temperature of the water as:

m_water * c_water * dT_water/dt = I^2 * R - k * A * (T_water - T_air)

where dT_water/dt is the rate of change of temperature of the water and I is the current passing through the resistor.

To solve this differential equation, we need to know the initial conditions (temperature of the water and surroundings) and the values of the constants (specific heat of water, exposed surface area, etc). Once we have this information, we can integrate the equation to obtain the temperature change over time.

As for extracting the specific heat of water from this setup, it is possible but it will require detailed information about the