What to solve for in specific heat energy problem

AI Thread Summary
The discussion centers on a calorimetric experiment aimed at measuring energy output from cells in relation to obesity research. Participants clarify that the goal is to compare energy input and output, using a resistor to determine power and measuring temperature changes in water. A key focus is identifying the heat transfer coefficient, which accounts for energy loss to the surroundings. The conversation also addresses the relationship between energy output, input, and the specific heat of water, emphasizing the need for accurate calibration. Participants seek guidance on correcting for energy loss during measurements to ensure precise results.
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Homework Statement


You have a position working with a group investigating biological mechanisms that
determine a predisposition to obesity. Your assignment is to measure the rate that
energy is output by certain types of cells when a nutrient is introduced. To begin this
study, you have decided to use a calorimetric technique. A culture of cells with the
appropriate nutrient is placed inside a closed container. That container is submerged in
a water bath and you measure the rate that the temperature of the bath changes.
To calibrate the apparatus, you decide to use a resistor connected across a known
voltage as a power source. You know that the power output by the resistor is just the
current through the resistor times the voltage across the resistor. You then will
compare that power to the rate that the internal energy of the water bath changes by
measuring its temperature as a function of time. To accomplish this calibration, you
calculate the rate of temperature change as a function of the voltage across the resistor,
the current through the resistor, the specific heat of the water, and the mass of the
water. You know that even with good insulation, your apparatus will transfer some
energy to the outside and your measurements will allow you to correct for this.


So I am honestly not sure what I am trying to solve for. I know that we are measuring energy output and that a resistor is used as a power source and I believe it is water we are measuring in terms of a transfer in energy from the system into the surronding environment. However, I'm confused as to what I'm specifically trying to solve for? What is it that I need to compare and look for in this problem? Any input is appreciated.
 
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The calorimeter loses energy by transferring heat to the surroundings at a rate proportional to the temperature difference between the calorimeter and the surroundings. You need to find that proportionality factor (heat transfer coefficient might be the name). Also, the calorimeter has its own heat capacity. When measuring temperature versus time with a given input power, you can determine both parameters.

ehild
 
So I believe I found a proportionality. Because the problem tells me that energy output can be determined by the power of the resistor and that energy input is determined as a function of temperature change over time of the water bath, energy input and energy output of the system are equivalent. I showed my work for this problem on the attached document. If someone could look at it to see if I'm on the right track with solving it, that would be appreciated.
 

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Just wanted to bump this up to see if anyone could look at my work from my previous post to the problem that I posted above. It'd be really appreciated, thanks!
 
I can not read docx files. Type in your solution.

ehild
 
Solution I came up with:



Efinal-Einitial = Eoutput-Einput
Efinal = 0
Einitial = 0
0 = Eoutput - Einput
Einput = Eoutput
Eoutput = Power of Resistor = Voltage (V) x Current (I)
dEoutput/dt = (mwater)(cwater)dT/dt
Power = (mwater)(cwater)dT/dt
Mwater = mass of water
Cwater = specific heat of water
∆Twater = change in temperature of water
 
To accomplish this calibration, you
calculate the rate of temperature change as a function of the voltage across the resistor,
the current through the resistor, the specific heat of the water, and the mass of the
water. You know that even with good insulation, your apparatus will transfer some
energy to the outside and your measurements will allow you to correct for this.

How do you do this correction for loss of energy? ehild
 
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