Temperature and Heat — Question 7

Section 18-6 · Physics pages 2924–2925

Question 7 A short contact pulse gives a temperature impulse A small sensor has constant heat capacity C = 4.00 J/K and initially reads 290 K. An ideal thermal pulse has rate P(t) = P0 sin(πt/τ) for 0 ≤t ≤τ, with P0 = 20.0 W and τ = 2.00 s. Neglect every other transfer and use C dT = P dt. t P A half-sine pulse lies below the peak-height rectangle. (a) Obtain T(t) during the pulse and the final temperature. (b) When has the sensor received half the total pulse heat? Compare the true heat with a rectangular-pulse estimate P0τ.
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Question 7 - Solution A short contact pulse gives a temperature impulse t P (a) Method and calculation T(t) = 290 + P0τ πC [1 −cos(πt/τ)]. Qtotal = 2P0τ π = 25.465 J, Tf = 296.366 K. (b) Interpretation and further calculation 1 −cos(πt/τ) = 1 ⇒t = τ/2 = 1.00 s. Qrectangle = 40.0 J, Qrectangle Qtotal = π 2 ≈1.571. The rectangle overestimates by 14.535 J, or about 57.1% of the true heat. Physical check The temperature rises monotonically because the input rate never becomes negative during the pulse.

Question 7

A short contact pulse gives a temperature impulse

A small sensor has constant heat capacity C=4.00J/KC=4.00\,\mathrm{J/K} and initially reads 290 K. An ideal thermal pulse has rate P(t)=P0sin⁡(πt/τ)P(t)=P_0\sin(\pi t/\tau) for 0≤t≤τ0\le t\le\tau, with P0=20.0WP_0=20.0\,\mathrm W and τ=2.00s\tau=2.00\,\mathrm s. Neglect every other transfer and use CdT=PdtC\,dT=P\,dt.

Diagram shown in the original SVG worksheet.

  1. Obtain T(t)T(t) during the pulse and the final temperature.

  2. When has the sensor received half the total pulse heat? Compare the true heat with a rectangular-pulse estimate P0τP_0\tau.

Question 7 - Solution

A short contact pulse gives a temperature impulse

Diagram shown in the original SVG worksheet.

(a) Method and calculation

T(t)=290+P0τπC[1−cos⁡(πt/τ)].T(t)=290+\frac{P_0\tau}{\pi C}[1-\cos(\pi t/\tau)].Qtotal=2P0τπ=25.465J,Tf=296.366K.Q_{\mathrm{total}}=\frac{2P_0\tau}{\pi}=25.465\,\mathrm J,\quad T_f=296.366\,\mathrm K.

(b) Interpretation and further calculation

1−cos⁡(πt/τ)=1⇒t=τ/2=1.00s.1-\cos(\pi t/\tau)=1\ \Rightarrow\ t=\tau/2=1.00\,\mathrm s.Qrectangle=40.0J,QrectangleQtotal=π2≈1.571.Q_{\mathrm{rectangle}}=40.0\,\mathrm J,\quad\frac{Q_{\mathrm{rectangle}}}{Q_{\mathrm{total}}}=\frac\pi2\approx1.571.The rectangle overestimates by 14.535 J, or about 57.1% of the true heat.

Physical check

The temperature rises monotonically because the input rate never becomes negative during the pulse.