Water is heated up with a small electric heater and the temperature of the water is monitored. The heater provides heat to the system with a thermal power PQ. The ontainer is a calorimeter that is assumed to absorb no heat. Before turning on the heater, the temperature of the water is T0 and its

Question

Water is heated up with a small electric heater and the temperature of the water is monitored. The heater provides heat to the system with a thermal power [latex]P_Q[/latex]. The container is a calorimeter that is assumed to absorb no heat. Before turning on the heater, the temperature of the water is [latex]T_0[/latex] and its entropy is [latex]S_0[/latex]. A linear temperature evolution given by [latex]T (t) = T_0 + At[/latex] is observed. Deduce the entropy variation ΔS during that process.

Accepted Answer

Since the system is simple, its evolution is reversible (2.23). The first law (2.22) implies
that,

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

[latex]ΠS = 0[/latex]. (2.23)

[latex]δQ=P_Qdt=T(S)d(S)[/latex].

[latex]\dot{S}=\frac{P_Q}{T(t)}=\frac{P_Q}{T_0+At}[/latex].

Thus, the entropy differential is expressed as,

[latex]dS=\frac{P_Q}{A} \Biggl(\frac{\frac{A}{T_0}dt }{1+\frac{A}{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}{\frac{P_Q}{A}\Biggl(\frac{\frac{A}{T_0}d\acute{t} }{1+\frac{A}{T_0}\acute{t} } \Biggr) } =S_0+\frac{P_Q}{A}\ln (1+\frac{A}{T_0}t)[/latex].

Thus, the entropy variation in water is given by,

[latex]ΔS=S(Δt)-S_0=\frac{P_Q}{A} \ln (1+\frac{A}{T_0}Δt)[/latex].

Question 2.11.2: Water is heated up with a small electric heater and the temp...

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