Temperature and Heat — Question 4

Section 18-6 · Physics pages 2918–2919

Question 4 Equal heat inputs need not cause equal temperature changes Two isolated test runs use bodies A and B, each initially at 300 K. Their constant heat capacities are CA = 200 J/K and CB = 500 J/K. With no work or phase change, use Q = C(Tf −Ti). Each receives 2.00 kJ. CA = 200 CB = 500 equal heat inputs Schematic; labels, not drawing lengths, specify the data. (a) Find both final temperatures and the ratio of their rises. (b) How much heat must each receive to reach 315 K? Explain why equal mass alone would not settle either comparison.
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Question 4 - Solution Equal heat inputs need not cause equal temperature changes CA = 200 CB = 500 equal heat inputs (a) Method and calculation ∆TA = 2000/200 = 10.0 K, ∆TB = 2000/500 = 4.00 K. Tf,A = 310 K, Tf,B = 304 K, ∆TA ∆TB = 2.50. (b) Interpretation and further calculation QA = 200(315 −300) = 3000 J. QB = 500(315 −300) = 7500 J. Heat capacity, not mass alone, fixes the response. Even at equal mass, differing material-specific capacities can give unequal C. Physical check For the same positive heat input, the larger heat capacity produces the smaller rise.

Question 4

Equal heat inputs need not cause equal temperature changes

Two isolated test runs use bodies A and B, each initially at 300 K. Their constant heat capacities are CA=200J/KC_A=200\,\mathrm{J/K} and CB=500J/KC_B=500\,\mathrm{J/K}. With no work or phase change, use Q=C(Tf−Ti)Q=C(T_f-T_i). Each receives 2.00 kJ.

Diagram shown in the original SVG worksheet.

  1. Find both final temperatures and the ratio of their rises.

  2. How much heat must each receive to reach 315 K? Explain why equal mass alone would not settle either comparison.

Question 4 - Solution

Equal heat inputs need not cause equal temperature changes

Diagram shown in the original SVG worksheet.

(a) Method and calculation

ΔTA=2000/200=10.0K,ΔTB=2000/500=4.00K.\Delta T_A=2000/200=10.0\,\mathrm K,\quad\Delta T_B=2000/500=4.00\,\mathrm K.Tf,A=310K,Tf,B=304K,ΔTAΔTB=2.50.T_{f,A}=310\,\mathrm K,\quad T_{f,B}=304\,\mathrm K,\quad\frac{\Delta T_A}{\Delta T_B}=2.50.

(b) Interpretation and further calculation

QA=200(315−300)=3000J.Q_A=200(315-300)=3000\,\mathrm J.QB=500(315−300)=7500J.Q_B=500(315-300)=7500\,\mathrm J.Heat capacity, not mass alone, fixes the response. Even at equal mass, differing material-specific capacities can give unequal CC.

Physical check

For the same positive heat input, the larger heat capacity produces the smaller rise.