Chap-6-HW-Solutions

Chap-6-HW-Solutions - Chapter 6 Homework Solutions 12. 13....

This preview shows pages 1–3. Sign up to view the full content.

Chapter 6 Homework Solutions 12. 454 g × 5.91 cal 1 g × 4.184 J 1 cal = 1.12 × 10 4 J 13 . (a) 297 kJ × 1000 J 1 kJ = 2.97 × 10 5 J (b) 2.97 × 10 5 J × 1 cal 4.184 J = 7.10 × 10 4 cal (c) 7.10 × 10 4 cal × 1 kcal 1000 cal = 71.0 kcal 24. Δ E water = q water + w water No work is done, so w water is zero: w water = 0 J Heat is transferred out of the system, so q sys is negative: q water = –840 J Δ E water = –840 J + (0 J) = –840 J Note, we did not need to use the sample mass or the change in temperature given, because the resulting heat was also given. 27. Table 6.1 tells us that the specific heat capacity (c) of Al is 0.902 J g –1 °C –1 and of liquid water is 4.184 J g –1 °C –1 . Equation 6.2: q = c × m × Δ T In both samples, the change in temperature is the same: Δ T = T f – T i = 37 °C – 25 °C = 12 °C (a) q water = (4.184 J g –1 °C –1 ) × (15.0 g) × (12 °C) = 750 J (b) q Al = (0.902 J g –1 °C –1 ) × (60.0 g) × (12 °C) = 650 J The energy required (750 J) to raise the temperature of the water sample is more than that required (650 J) to raise the temperature of the aluminum sample. 29. Table 6.1 tells us that the specific heat capacity (c) of Al is 0.902 J g –1 °C –1 . Equation 6.2: q = c × m × Δ T Δ T = T f – T i = 250 °C – 25 °C = 225 °C 230 °C (must be rounded to tens place) q Al = (0.902 J g –1 °C –1 ) × (500. g) × (230 °C) = 1.0 × 10 5 J 35 . (a) Equation 6.2: q = c × m × Δ T

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
c = q m ×Δ T = 41.0 J 12.3 g × 24.7 o C 17.3 o C = 41.0 J 12.3 g × 7.4 o C = 0.45 J g o C (b) The molar mass of iron is 55.85 g/mol. Use that to calculate the molar heat capacity: 0.45 J g o C × 55.845 g 1 mol = 25 J mol o C 41 . The thermal energy enters the surroundings as heat, so the temperature in the surroundings, T surroundings , rises: T f,surroundings > T i,surroundings Δ T surroundings = T f,surroundings – T i,surroundings = positive Here, the surroundings gets energy from the system, so the internal energy of the system, E system , lowers: E f,system < E i,system Δ E system = E f,system – E i,system = negative 42. (a) The thermal energy from the surroundings is used to cause the physical phase transition, so: T f,alloy = T i,alloy Δ T alloy = T f,alloy – T i,alloy = zero (neither positive nor negative) (b) Here, the system gets energy from the surroundings, so the internal energy of the system, E alloy , rises. E f,alloy > E i,alloy Δ E alloy = E f,alloy – E i,alloy = positive 46. Define the system as the H 2 O. m ice = (4 cubes) × 60.1 g H 2 O cube = 240. g = m water = m vapor To change solid to liquid in the system, use q = m × Δ H fus . q
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 09/13/2008 for the course CHEM 1410 taught by Professor Acree during the Spring '08 term at North Texas.

Page1 / 7

Chap-6-HW-Solutions - Chapter 6 Homework Solutions 12. 13....

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online