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I heated up a beaker of water to 70˚C and set a time of 2 minutes. I measured the change in temperature in that time interval. I changed the mass of water and measured the change in temperature each time. The surface area of the water and the temperature of the surroundings were constant.

I am wondering if I can use the equation for specific heat capacity, but am not sure how the thermal energy variable would play into this. Is there any other equation I can use?

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the simplest equation is (heat added to an object) = (mass of object) x (specific heat capacity) x (change in temperature, degrees absolute). This works when there are no heat losses, phase changes, or work extracted during the heat addition process.

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  • $\begingroup$ Why does it habve to be change in absolute temperature? $\endgroup$ – Chet Miller Dec 26 '17 at 1:45
  • $\begingroup$ that is just a habit of mine, it helps me avoid mistakes where temperatures are not expressed as differences and must be in degrees absolute. $\endgroup$ – niels nielsen Dec 26 '17 at 2:27
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Cooling of objects is a complicated business. In your experiment it involves the flow of air around the beaker, the circulation of water in the beaker and heat transport through the glass walls of the beaker.

However in many cases the cooling can be described by an approximate law called Newton's law of cooling. This tells us that the rate of heat loss by the cooling object is proportional to the temperature difference between the object and its environment:

$$ \frac{dQ}{dt} = k\left(T_\text{obj} - T_\text{env}\right) $$

where $T_\text{obj}$ is the temperature of the object and $T_\text{env}$ is the temperature of the environment. Now if we lose some amount of heat $dQ$ the temperature change $dT$ is given by:

$$ dQ = -CdT $$

where $C$ is the specific heat of our object. Substituting in our differential equation gives us:

$$ \frac{dT_\text{obj}}{dt} = -\frac{k}{C}\left(T_\text{obj} - T_\text{env}\right) $$

For convenience we tend to define $\Delta T = T_\text{obj} - T_\text{env}$ in which case we can write:

$$ \frac{d\Delta T}{dt} = -\frac{k}{C}\Delta T $$

And solving this differential equation gives us the equation the temperature difference as a function of time:

$$ \Delta T(t) = \Delta T_0 e^{-(k/C)t} $$

And this is the equation that describes the temperature change of your object. For experimental purposes it's probably more convenient to write:

$$ \ln\left(\frac{\Delta T}{\Delta T_0}\right) = -\frac{k}{C} t $$

because if you draw a graph of $\ln(\Delta T/\Delta T_0)$ against time you should get a straight line of gradient $k/C$. So that draw this graph for your different amounts of water and calculate $k/C$.

The trouble is that the constant $k$ is a fudge factor that groups together all the factors like air flow, water flow in the beaker, transport through the glass walls, etc into one rather ill defined constant.

The specific heat $C$ is well defined, because it is just the mass of your water multiplied by the specific heat of water $4.184$ J/kg/K. So if you double the amount of water you will double the value of $C$ and if $k$ is constant this will halve the gradient of your graph. However I suspect that $k$ will be changing as well. For example the area through which heat is flowing is included in the value of $k$, and this area will change as the amount of water changes.

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