In an experiment similar to the Joule experiment (Fig. 1.1), an electric motor is used instead of a weight to stir the liquid. The thermal power PQ, assumed to result from the friction, is known. The coefficient cM, which represents the heat per unit mass and temperature, is known and it is assumed

Question

In an experiment similar to the Joule experiment (Fig. 1.1), an electric motor is used instead of a weight to stir the liquid. The thermal power [latex]P_Q[/latex], assumed to result from the friction, is known. The coefficient [latex]c_M[/latex], which represents the heat per unit mass and temperature, is known and it is assumed to be independent of temperature.
a) Deduce the temperature rise ΔT after stirring for a time Δt.
b) Find an expression for the entropy variation ΔS during this process, which started at temperature [latex]T_0[/latex].
Numerical Application:
M = 200 g, [latex]P_Q = 19 W, c_M = 3 J g^{−1}K^{−1}[/latex], Δt = 120 s, [latex]T_0 = 300[/latex] K.

Accepted Answer

a) According to the first law, the increase of internal energy ΔU during a time interval Δt is given by,

[latex]ΔU = P_Q Δt[/latex].

The increase of internal energy ΔU for a temperature increase ΔT of the liquid yields,

[latex]ΔU = Mc_M ΔT[/latex].

Thus, by comparing these two equations, the temperature increase ΔT is found to be,

[latex]ΔT =\frac{P_Q Δt}{Mc_M}= 3.8K[/latex].

b) The first law (2.22) is written as,

[latex]\dot{U}(S) = T(S) \dot{U} = P_Q[/latex]. (2.22)

[latex]\dot{U} = T\dot{S} = P_Q[/latex] thus [latex]\dot{S} = \frac{P_Q }{T}[/latex].

According to a) the temperature is given by,

[latex]T=T_0+\frac{P_Q}{Mc_M} t[/latex].

which implies that,

[latex] dS =\frac{P_Q}{T} dt= \frac{P_Q}{T_0+\frac{P_Q}{Mc_M}t } =Mc_M \Biggl(\frac{\frac{P_Q}{Mc_M T_0}dt }{1+\frac{P_Q}{Mc_M T_0}t }\Biggr) [/latex].

When integrating this equation over time, we obtain,

[latex]S(t) = \int_{S_0}^{S(t)}{dS} = \int_{0}^{t}{Mc_M} \Biggl(\frac{\frac{P_Q}{Mc_M T_0}dt^{\prime} }{1+\frac{P_Q}{Mc_M T_0}t^{\prime} }\Biggr)[/latex]

[latex]= S_0 + Mc_M \ln \Bigl(1+\frac{P_Q t}{Mc_M T_0} \Bigr)[/latex].

Thus, the entropy increase during the stirring process is,

[latex]ΔS = S (Δt) − S_0 = Mc_M \ln \Bigl(1+\frac{P_Q \Delta t}{Mc_M T_0} \Bigr)[/latex].

Question 2.5: In an experiment similar to the Joule experiment (Fig. 1.1),...

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