Temperature vs Heat: Double the Temp, Double the Heat?

[Jonnyb42]

In terms of Kelvins, if the temperature of a sample doubles, does the energy content from heat double as well?

Also, if you are passing a current through a sample, what should be the temperature vs current relation be theoretically? i.e. what is f(I) in T = f(I).

Thanks,

 

[Ken G]

In terms of Kelvins, if the temperature of a sample doubles, does the energy content from heat double as well?

Usually, though not always. It's not a fundamental feature of the connection between energy and temperature, but it is a common connection.

Also, if you are passing a current through a sample, what should be the temperature vs current relation be theoretically?

Generally one talks about the heating rate generated by a current, according to P = I²/R. So given I and R, you can find the heating rate. What that does to T depends on the cooling rate. Sometimes the cooling rate is proportional to T, so in equilibrium T would be proportional to I², but there are many other ways that the cooling could depend on T.

 

[Andrew Mason]

In terms of Kelvins, if the temperature of a sample doubles, does the energy content from heat double as well?

Just to add to what Ken G has said, temperature is a measure of the translational kinetic energy of the molecules in the sample. But the internal energy of a sample is not just the translational kinetic energy. It can include potential energy as well. A good example is water. It takes a lot of energy to turn ice at 0C into water at 0C. This is the energy required just to break enough of the inter-molecular bonds between water molecules to turn the substance from a solid to a liquid. The liquid water molecules at 0C have much more potential energy than ice - 334 J per gram more energy - but temperature has not increased at all.

Also, if you are passing a current through a sample, what should be the temperature vs current relation be theoretically? i.e. what is f(I) in T = f(I).

Use the first law of thermodynamics: Q = ΔU + W.

Lets assume you pass a current I through a resistance R for one second and no mechanical work (such as turning a motor to lift something) is done (W = 0). In that case, Q = ΔU. The electrical energy flow is heat flow: Q = I²RΔt = I²R (since Δt = 1) The units would be in Joules.

In general if no mechancal work is done, in time t: Q = ΔU = I²Rt.

Assuming a constant heat capacity for the sample over the temperature change that this energy causes, then Q = mC_vΔT where m is the mass and C_v is the heat capacity at constant volume. So ΔT = I²Rt/mC_v.

AM