Temperature — Question 9

Question 9

A finite probe changes the sample it measures

A sample initially at unknown $T_s$ contacts a probe initially at $280\,\mathrm K$. Their thermal parameters are $C_s=40\,\mathrm{J/K}$ and $C_p=10\,\mathrm{J/K}$. The isolated pair reaches $T_f=304\,\mathrm K$. Use the supplied balance $C_s(T_f-T_s)+C_p(T_f-T_p)=0$, with constant parameters and no phase change.

1. Infer the original sample temperature and the error of calling $T_f$ its original value.

2. For a sample initially at $310\,\mathrm K$ and a probe at $280\,\mathrm K$, determine the largest $C_p/C_s$ that keeps the disturbance no greater than $0.50\,\mathrm K$.

Question 9 - Solution

A finite probe changes the sample it measures

(a) Method and calculation

$$T_s=T_f+\frac{C_p}{C_s}(T_f-T_p)=304+\frac{10}{40}(24)=310\,\mathrm K.$$$$T_f-T_s=-6.0\,\mathrm K.$$The equilibrium reading belongs to the final combined system.

(b) Interpretation and further calculation

Let $r=C_p/C_s\ge0$. Rearranging the supplied balance gives$$T_f=\frac{T_s+rT_p}{1+r},\qquad T_s-T_f=\frac{30r}{1+r}\,\mathrm K.$$$$\frac{30r}{1+r}\le.50\ \Longrightarrow\ r\le\frac1{59}\approx.01695.$$A sufficiently small probe reduces disturbance; fast response alone does not guarantee this.

Physical check

The inferred $310$ K exceeds the final $304$ K because the initially cooler probe warmed at the sample’s expense.