{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

CH302 Homework 2-solutions

# CH302 Homework 2-solutions - Bodet(ngb299 – Homework 2...

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

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.

Unformatted text preview: Bodet (ngb299) – Homework 2 – sutcliffe – (51630) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. NOTE: AFTER QUESTION 15 READ CAREFULLY! Some questions will require you to consider the vant’Hoff factor, i. If the solute is stated to be a nonelectrolyte, or if it is a sugar, i = 1. Assume complete dissocia- tion of ionic compounds. 001 10.0 points The molar heat of vaporization of carbon disulfide (CS 2 ) is 28.4 kJ/mol at its nor- mal boiling point of 46 ◦ C. How much energy (heat) is required to vaporize 5 . 7 g of CS 2 at 46 ◦ C? Correct answer: 2126 . 08 Joules. Explanation: 002 10.0 points How much heat energy is needed to raise the temperature of a 50 g sample of aluminum from 35 . 2 ◦ C to 90 . 5 ◦ C? The specific heat of aluminum is 0 . 897 J / g · K. Correct answer: 2480 . 2 J. Explanation: m = 50 g SH Al = 0 . 897 J / g · K Δ T = 90 . 5 ◦ C- 35 . 2 ◦ C = 55 . 3 ◦ C = 55 . 3 K q = ? SH Al = q m Δ T q = SH Al m Δ T = (0 . 897 J / g · K) (50 g) (55 . 3 K) = 2480 . 2 J 003 10.0 points Consider the following specific heats SH H 2 O(s) = 2 . 09 J / g · ◦ C , SH H 2 O( ) = 4 . 18 J / g · ◦ C , and SH H 2 O(g) = 2 . 03 J / g · ◦ C. The heat of fusion for water is 334 J/g and the heat of vaporization for water is 2260 J/g. Calculate the amount of heat required to convert 51 g of ice at- 37 ◦ C completely to steam at 131 ◦ C. Correct answer: 160 . 765 kJ. Explanation: SH H 2 O(s) = 2 . 09 J / g · ◦ C Δ H vap = 2260 J/g SH H 2 O( ) = 4 . 18 J / g · ◦ C Δ H fus = 334 J/g SH H 2 O(g) = 2 . 03 J / g · ◦ C m ice = 51 g T 1 =- 37 ◦ C T 2 = 131 ◦ C In this example, ice is converted to steam. Five separate steps take place here: 1) ice is warmed from- 37 ◦ C to 0 ◦ C; 2) a phase change from ice to water at 0 ◦ C; 3) water is warmed from 0 ◦ C to 100 ◦ C; 4) a phase change from water to gas at 100 ◦ C; 5) steam is warmed from 100 ◦ C to 131 ◦ C. Each of these steps involves a separate en- ergy calculation, and the sum of the energies of these five steps is the total amount of heat required for the process. For processes that involve warming a solid, liquid, or gas, we use the equation q = (SH) m Δ T , where SH = specific heat, m = mass or moles, and Δ T = temperature change. For phase changes (solid to liquid or liquid to gas), we use q = Δ H vap or fus (m) , where again m = mass or moles. So for the first step, warming ice, we have q 1 = (51 g) 2 . 09 J g · ◦ C ◦ C- (- 37 ◦ C) = 3940 J . For the second step, the phase change from ice to liquid: q 2 = (334 J / g)(51 g) = 17000 J . For the third step, warming water: q 3 = (51 g) 4 . 18 J g · ◦ C (100 ◦ C- ◦ C) = 21300 J . Bodet (ngb299) – Homework 2 – sutcliffe – (51630) 2 For the fourth step, the phase change from liquid to gas: q 4 = (2260 J / g)(51 g) = 115000 J ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 7

CH302 Homework 2-solutions - Bodet(ngb299 – Homework 2...

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

View Full Document
Ask a homework question - tutors are online