Temperature and Heat — Question 8

Section 18-6 · Physics pages 2926–2927

Question 8 A temperature plateau still admits heat transfer A pure sample is at its melting temperature with solid and liquid coexisting. In the ideal model its temperature remains fixed while melting requires latent energy L = 200 kJ/kg. Initially it contains .30 kg solid and .20 kg liquid. Heat enters at 50 W; neglect other transfers. solid + liquid 50 W Schematic; labels, not drawing lengths, specify the data. (a) Find heat absorbed and solid remaining after 400 s. (b) Find the time at which melting just finishes. Can the subsequent temperature be found from L alone?
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Question 8 - Solution A temperature plateau still admits heat transfer solid + liquid 50 W (a) Method and calculation Q = Pt = 50(400) = 20000 J. mmelted = Q/L = .100 kg, msolid = .300 −.100 = .200 kg. Temperature stays at the melting value while both phases remain. (b) Interpretation and further calculation Qfinish = .300(200000) = 60000 J. tfinish = 60000/50 = 1200 s. Afterward one needs the liquid heat capacity, and any relevant loss law, to determine its temperature. The latent-energy parameter does not describe warming of a single phase. Physical check ∆T = 0 is compatible with Q > 0 when the phase fraction changes.

Question 8

A temperature plateau still admits heat transfer

A pure sample is at its melting temperature with solid and liquid coexisting. In the ideal model its temperature remains fixed while melting requires latent energy L=200kJ/kgL=200\,\mathrm{kJ/kg}. Initially it contains .30 kg solid and .20 kg liquid. Heat enters at 50 W; neglect other transfers.

Diagram shown in the original SVG worksheet.

  1. Find heat absorbed and solid remaining after 400 s.

  2. Find the time at which melting just finishes. Can the subsequent temperature be found from LL alone?

Question 8 - Solution

A temperature plateau still admits heat transfer

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Q=Pt=50(400)=20000J.Q=P t=50(400)=20000\,\mathrm J.mmelted=Q/L=.100kg,msolid=.300−.100=.200kg.m_{\mathrm{melted}}=Q/L=.100\,\mathrm{kg},\quad m_{\mathrm{solid}}=.300-.100=.200\,\mathrm{kg}.Temperature stays at the melting value while both phases remain.

(b) Interpretation and further calculation

Qfinish=.300(200000)=60000J.Q_{\mathrm{finish}}=.300(200000)=60000\,\mathrm J.tfinish=60000/50=1200s.t_{\mathrm{finish}}=60000/50=1200\,\mathrm s.Afterward one needs the liquid heat capacity, and any relevant loss law, to determine its temperature. The latent-energy parameter does not describe warming of a single phase.

Physical check

ΔT=0\Delta T=0 is compatible with Q>0Q>0 when the phase fraction changes.