Chapter 7 Solutions

# Chapter 7 Solutions - Chapter 7 Answers 7 Heat is flowing...

This preview shows pages 1–4. Sign up to view the full content.

Chapter 7 Answers 7) Heat is flowing from the block of Mg to the water, until they equilibrate, that is. We will be using the relationship: q Mg =-q H2O mc Mg T=-mc H2O T For Mg: m=1.00Kg = 1000g c=1.024J g -1 ° C -1 T i =40.0 ° C T F =? For water: m=1.00Lx(1000mL/1L)x(1g/mL)=1000g c=4.18 J g -1 ° C -1 (this is a value you are expected to know from reading the chapter) T i =20.0 ° C T F =? It is important to note that, at equilibrium, Tf for water and Tf for Mg will be the same. Now we just substitute these values into our equation and solve for Tf to get: mc Mg T=-mc H2O T 1000g(1.024)(Tf-40)= - 1000g(4.18)(Tf-20) Tf=24.0 ° C *note* I am lazy with units because they tend to make the typed equations more difficult to follow I will provide the correct units for the variables list, and the correct units for the answer. I strongly suggest working through the units on your own as an exercise so that you can see where each unit comes from. 9) This is similar to the previous question, except we are solving for c glycerol . q Cu =-q gly mc Cu T=-mc Gly T For Cu: m=74.8g c=0.385J g -1 ° C -1 T i =143.2 ° C T F =31.1 ° C

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

View Full Document
For glycerol: m=165mLx(1.26g/mL)=208g c=? J g -1 ° C -1 T i =24.8 ° C T F =31.1 ° C Mc Cu T=-mc Gly T 74.8g(0.385)(31.1-143.2)= - 208g(c)(31.1-24.8) C=2.46 J g -1 ° C -1 Molar heat capacity is found by multiplying by the molecular weight of glycerol (MW=92.1g/mol). Molar heat capacity=2.46 J g -1 ° C -1 x 92.1g/mol = 236 J mol -1 ° C -1 10) This one is hard to visualize at first. What you have to remember is that the water already in the calorimeter must remain at a constant temperature. Therefore, q=0 for this water (Tf-Ti=0). Now, we can focus on the two systems that actually change: the iron and the ‘added water’. q Fe =-q H2O mc Fe T=-mc H2O T For Fe: m=1.23Kg = 1230g c=0.449 J g -1 ° C -1 T i =68.5 ° C T F =25.6 ° C For water: m=? c=4.18 J g -1 ° C -1 (this is a value you are expected to know from reading the chapter) T i =18.5 ° C T F =25.6 ° C mc Fe T=-mc H2O T 1230g(0.449)(25.6-68.5)= -m (4.18)(25.6-18.5) m=800g However, it is important to note that the question asks for the volume of water, so we need to divide by the density (density=mass/volume). The density of water is 1g/mL, so 800g=800mL 17) According to the balanced reaction, each mole of methane releases 890.3 KJ of gas. a) With this knowledge, we can determine how many moles of methane yield 2.8x10 7 kJ: 2.8x10 7 kJ/890.3kJ mol -1 = 31450moles
Molecular weight = mass/moles Mass=molesxMW =31450moelsx16g/mol =503.2Kg b) Ideal gas law! PV=nRT

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

### Page1 / 9

Chapter 7 Solutions - Chapter 7 Answers 7 Heat is flowing...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online