The Absorption of Heat by Solids and Liquids — Question 9

Section 18-7 · Physics pages 2948–2949

Question 9 An incomplete-melting measurement determines latent heat A .100 kg solid sample is initially at its melting point. Its vessel has negligible heat capacity. During a 100 s interval a heater supplies 250 W, while a measured constant 25 W leaks to the environment. At the end, .0750 kg has melted, with both phases still present. melting sample loss Schematic; labels, not drawing lengths, specify the data. (a) Find the latent heat and the value inferred if the leak were ignored. (b) At the same powers, calculate the additional time required to melt the remaining solid and the total energy lost to the environment over the entire melting run.
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Question 9 - Solution An incomplete-melting measurement determines latent heat melting sample loss (a) Method and calculation Qsample = (250 −25)(100) = 22500 J. Lf = 22500/.0750 = 300000 J/kg. Lignore = 25000/.0750 = 333333 J/kg, an 11.11% overestimate. (b) Interpretation and further calculation tadditional = (.100 −.0750)(300000) 225 = 33.3333 s. Qleak,total = 25(100 + 33.3333) = 3333.33 J. The heater supplies 33333.33 J in total, of which 30000 J melts the sample. Physical check The plateau justifies latent-heat accounting only while some solid remains.

Question 9

An incomplete-melting measurement determines latent heat

A .100 kg solid sample is initially at its melting point. Its vessel has negligible heat capacity. During a 100 s interval a heater supplies 250 W, while a measured constant 25 W leaks to the environment. At the end, .0750 kg has melted, with both phases still present.

Diagram shown in the original SVG worksheet.

  1. Find the latent heat and the value inferred if the leak were ignored.

  2. At the same powers, calculate the additional time required to melt the remaining solid and the total energy lost to the environment over the entire melting run.

Question 9 - Solution

An incomplete-melting measurement determines latent heat

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Qsample=(250−25)(100)=22500J.Q_{\mathrm{sample}}=(250-25)(100)=22500\,\mathrm J.Lf=22500/.0750=300000J/kg.L_f=22500/.0750=300000\,\mathrm{J/kg}.Lignore=25000/.0750=333333J/kg,L_{\mathrm{ignore}}=25000/.0750=333333\,\mathrm{J/kg},an 11.11% overestimate.

(b) Interpretation and further calculation

tadditional=(.100−.0750)(300000)225=33.3333s.t_{\mathrm{additional}}=\frac{(.100-.0750)(300000)}{225}=33.3333\,\mathrm s.Qleak,total=25(100+33.3333)=3333.33J.Q_{\mathrm{leak,total}}=25(100+33.3333)=3333.33\,\mathrm J.The heater supplies 33333.33 J in total, of which 30000 J melts the sample.

Physical check

The plateau justifies latent-heat accounting only while some solid remains.