# Liquids, internal energy and specific heat capacity

A liquid contained in an adiabatic container is shaked vigorously so that it its temp. Increases.

The heat capacity for the liquid is given, the rise in temp. Is given.

According to the first law of thermo, dQ=dW + dU
here dQ is 0.

Asked, is to find the work done on the system, i.e. -dW =dU

The correct solution to this is dU = m*s*dT

My question is , how can this formula be used ?? Since the correct formula is dQ=m*s*dT
??

No the correct formula is not q = msΔT.

q is the heat transferred into the system and this is defined to be zero by the adiabatic nature of the process.

Work is transferred into the system.
Within the system this work is converted to heat by interparticulate friction.
This heat increases the temperature of the system.

The correct equation connecting tmeperature rise to heat used to effect that temperature rise is

Q = mass times specific heat times temperature rise.

Q is numerically (and factually) equal to the work transferred into the system.

You should remember that q and w are always the energy that crosses the boundary into or out of the system when used in the first law.
The internal energy is a sort of book keeping account to balance the inputs and outputs.

Chestermiller
Mentor
A liquid contained in an adiabatic container is shaked vigorously so that it its temp. Increases.

The heat capacity for the liquid is given, the rise in temp. Is given.

According to the first law of thermo, dQ=dW + dU
here dQ is 0.

Asked, is to find the work done on the system, i.e. -dW =dU

The correct solution to this is dU = m*s*dT

My question is , how can this formula be used ?? Since the correct formula is dQ=m*s*dT
??
Actually, the more basic (and correct) formula for an incompressible liquid is dU = m*s*dT. If you add heat to an incompressible liquid that is not being deformed, you get dU = m*s*dT = dQ, and this gives you a relationship between the amount of heat added to the temperature change. On the other hand, if you do work to deform an incompressible liquid without adding heat, you get dQ = 0, but still have dU = m*s*dT.

Actually, the more basic (and correct) formula for an incompressible liquid is dU = m*s*dT. If you add heat to an incompressible liquid that is not being deformed, you get dU = m*s*dT = dQ, and this gives you a relationship between the amount of heat added to the temperature change. On the other hand, if you do work to deform an incompressible liquid without adding heat, you get dQ = 0, but still have dU = m*s*dT.
I think that would be correct. I was thinking along the same lines initially. But the standard textbook definition of heat capacities involve heat, i.e. Q and not U. Are you sure your formulation is correct?

And if it is, could you point me to some textbooks or resources where I could find heat capacities defined in terms of dU ?

Chestermiller
Mentor
I think that would be correct. I was thinking along the same lines initially. But the standard textbook definition of heat capacities involve heat, i.e. Q and not U. Are you sure your formulation is correct?

And if it is, could you point me to some textbooks or resources where I could find heat capacities defined in terms of dU ?
Yes. Defining it in terms of Q is just for beginners.

See Amazon under the key word Thermodynamics

Smith and Van Ness

Van Ness and Abbot