The Absorption of Heat by Solids and Liquids — Question 6

Section 18-7 · Physics pages 2942–2943

Question 6 Condensing steam supplies two distinct energy contributions Steam at 100◦C is injected into .400 kg water at 20.0◦C in an insulated cup of negligible heat capacity. All injected steam condenses, and the final temperature is 60.0◦C. Use Lv = 2.26 × 106 J/kg and cw = 4200 J/(kg K). Ignore steam kinetic energy and losses. water at 60◦C steam Schematic; labels, not drawing lengths, specify the data. (a) Find the required injected steam mass, including cooling of the condensate. (b) Find the result if condensate cooling is omitted and the resulting percentage overestimate of steam mass.
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Question 6 - Solution Condensing steam supplies two distinct energy contributions water at 60◦C steam (a) Method and calculation ms[Lv + cw(100 −60)] = .400cw(60 −20). ms = 67200 2260000 + 168000 = .0276771 kg. Each kilogram releases latent heat and then sensible heat as the condensate cools. (b) Interpretation and further calculation ms,latent only = 67200/2260000 = .0297345 kg. ms,latent only −ms ms × 100% = 168000 2260000 × 100% ≈7.4336%. Omitting a real energy contribution overestimates the required source mass. Physical check The final temperature is below the condensation temperature, consistent with fully liquid retained water.

Question 6

Condensing steam supplies two distinct energy contributions

Steam at 100∘100^\circC is injected into .400 kg water at 20.0∘20.0^\circC in an insulated cup of negligible heat capacity. All injected steam condenses, and the final temperature is 60.0∘60.0^\circC. Use Lv=2.26×106J/kgL_v=2.26\times10^6\,\mathrm{J/kg} and cw=4200J/(kgK)c_w=4200\,\mathrm{J/(kg\,K)}. Ignore steam kinetic energy and losses.

Diagram shown in the original SVG worksheet.

  1. Find the required injected steam mass, including cooling of the condensate.

  2. Find the result if condensate cooling is omitted and the resulting percentage overestimate of steam mass.

Question 6 - Solution

Condensing steam supplies two distinct energy contributions

Diagram shown in the original SVG worksheet.

(a) Method and calculation

ms[Lv+cw(100−60)]=.400cw(60−20).m_s[L_v+c_w(100-60)]=.400c_w(60-20).ms=672002260000+168000=.0276771kg.m_s=\frac{67200}{2260000+168000}=.0276771\,\mathrm{kg}.Each kilogram releases latent heat and then sensible heat as the condensate cools.

(b) Interpretation and further calculation

ms,latentonly=67200/2260000=.0297345kg.m_{s,\mathrm{latent\ only}}=67200/2260000=.0297345\,\mathrm{kg}.ms,latentonly−msms×100%=1680002260000×100%≈7.4336%.\frac{m_{s,\mathrm{latent\ only}}-m_s}{m_s}\times100\%=\frac{168000}{2260000}\times100\%\approx7.4336\%.Omitting a real energy contribution overestimates the required source mass.

Physical check

The final temperature is below the condensation temperature, consistent with fully liquid retained water.