Concept explainers
(a)
Interpretation:
The amount of heat energy needed to melt 45 g of ice has to be given.
Concept introduction:
The specific heat is the amount of heat energy needed to raise the temperature of 1g of a substance by 1oC.
Specific heat=Heatmass×ΔT
(a)
Answer to Problem 4.75AP
The amount of energy needed to melt ice is 2475 cal.
Explanation of Solution
The amount of energy needed to melt ice:
Given,
The mass of ice melt is 45 g.
The specific heat of water is 1.00 cal/g oC.
The temperature at initial (T1) is 0oC.
The temperature at final (T2) is 55oC.
The change in temperature (ΔT) can be calculated as,
ΔT=T2-T1
ΔT=55oC
The amount of energy needed to melt ice can be given as,
Specific heat=Heatmass×ΔT
Heat = mass ×ΔT × Specific heat
Heat = 45 g×55oC × 1.00 cal/goC
Heat = 2475 cal
The amount of energy needed to melt ice is 2475 cal.
(b)
Interpretation:
The amount of heat energy released when 45 g of water is cooled has to be given.
Concept introduction:
Refer to part (a).
(b)
Answer to Problem 4.75AP
The amount of energy released when water is cooled is − 2475 cal.
The amount of heat energy released when 45 g of water frozen is 3587 cal.
Explanation of Solution
The amount of heat energy released when 45 g of water is cooled:
Given,
The mass of water is 45 g.
The specific heat of water is 1.00 cal/g oC.
The temperature at initial (T1) is 55oC.
The temperature at final (T2) is 0oC.
The change in temperature (ΔT) can be calculated as,
ΔT=T2-T1
ΔT= -55oC
The amount of energy released when water is cooled can be given as,
Specific heat=Heatmass×ΔT
Heat = mass ×ΔT × Specific heat
Heat = 45 g×- 55oC × 1.00 cal/goC
Heat = −2475 cal
The amount of energy released when water is cooled is − 2475 cal.
The amount of energy released when water if frozen to ice can be calculated as,
The mass of water is 45 g.
The heat of fusion of water is 79.7 cal/g.
The specific heat of water is 1.00 cal/g oC.
The amount of energy needed to melt ice can be given as,
mass of substance(g)×heat of fusion(cal)1g
45 g×79.7 cal1g=3587 calories
The amount of heat energy released when 45 g of water is frozen is 3587 cal.
(c)
Interpretation:
The amount of energy released when 35 g of steam condensed has to be given.
Concept introduction:
Refer to part (a).
(c)
Answer to Problem 4.75AP
The amount of energy released when water is condensed is - 3.500 kcal.
The amount of heat energy released when 35 g of water is cooled is 2.79 kcal.
Explanation of Solution
The amount of energy when steam condensed:
Given,
The mass of steam is 35 g.
The heat of vaporization of water is 540 cal/g.
The specific heat of water is 1.00 cal/g oC.
The heat of fusion of water is 79.7 cal/g.
The temperature at initial (T1) is 100oC.
The temperature at final (T2) is 0oC.
The change in temperature (ΔT) can be calculated as,
ΔT=T2-T1
ΔT= -100oC
The amount of energy released when water is condensed can be given as,
Specific heat=Heatmass×ΔT
Heat = mass ×ΔT × Specific heat
Heat = 35 g×- 100oC × 1.00 cal/goC
Heat = −3500 cal
The energy in cal can be converted to kcal as,
1 kcal=1,000 cal
-3500 cal×1 kcal1000 cal= -3.500 kcal
The amount of energy released when water is condensed is - 3.500 Kcal.
The amount of energy released when water if frozen to ice can be calculated as,
The amount of energy needed to froze can be given as,
mass of substance(g)×heat of fusion(cal)1g
35 g×79.7 cal1g=3347 calories
The amount of heat energy released when 35 g of water is frozen is 2790 cal.
The energy in cal can be converted to Kcal as,
1 kcal=1,000 cal
2790 cal×1 kcal1000 cal=2.79 kcal
The amount of heat energy released when 35 g of water is frozen is 2.79 kcal.
Want to see more full solutions like this?
Chapter 4 Solutions
Principles of General, Organic, Biological Chemistry
- Don't used hand raitingarrow_forwardShown below is the major resonance structure for a molecule. Draw the second best resonance structure of the molecule. Include all non-zero formal charges. H. H. +N=C H H H Cl: Click and drag to start drawing a structure. : ? g B S olo Ar B Karrow_forwardDon't used hand raitingarrow_forward
- S Shown below is the major resonance structure for a molecule. Draw the second best resonance structure of the molecule. Include all non-zero formal charges. H H = HIN: H C. :0 H /\ H H Click and drag to start drawing a structure. ×arrow_forwardPlease help me figure out these calculation and what should be plotted. These are notes for my chemistry class.arrow_forwardNonearrow_forward
- Nonearrow_forwardPart II. two unbranched ketone have molecular formulla (C8H100). El-ms showed that both of them have a molecular ion peak at m/2 =128. However ketone (A) has a fragment peak at m/2 = 99 and 72 while ketone (B) snowed a fragment peak at m/2 = 113 and 58. 9) Propose the most plausible structures for both ketones b) Explain how you arrived at your conclusion by drawing the Structures of the distinguishing fragments for each ketone, including their fragmentation mechanisms.arrow_forwardPart V. Draw the structure of compound tecla using the IR spectrum Cobtained from the compound in KBr pellet) and the mass spectrum as shown below. The mass spectrum of compound Tesla showed strong mt peak at 71. TRANSMITTANCE LOD Relative Intensity 100 MS-NW-1539 40 20 80 T 44 55 10 15 20 25 30 35 40 45 50 55 60 65 70 75 m/z D 4000 3000 2000 1500 1000 500 HAVENUMBERI-11arrow_forward
- Technetium is the first element in the periodic chart that does not have any stable isotopes. Technetium-99m is an especially interesting and valuable isotope as it emits a gamma ray with a half life ideally suited for medical tests. It would seem that the decay of technetium should fit the treatment above with the result In(c/c) = -kt. The table below includes data from the two sites: http://dailymed.nlm.nih.gov/dailymed/druginfo.cfm?id=7130 http://wiki.medpedia.com/Clinical: Neutrospec_(Technetium_(99m Tc)_fanolesomab). a. b. C. Graph the fraction (c/c.) on the vertical axis versus the time on the horizontal axis. Also graph In(c/c.) on the vertical axis versus time on the horizontal axis. When half of the original amount of starting material has hours fraction remaining disappeared, c/c = ½ and the equation In(c/c.) = -kt becomes In(0.5) = -kt1/2 where t₁₂ is the half life (the time for half of the material to decay away). Determine the slope of your In(c/c.) vs t graph and…arrow_forwardPlease correct answer and don't use hand ratingarrow_forward1. a) Assuming that an atom of arsenic has hydrogen-like atomic orbitals, sketch the radial probability plots for 4p and 4d orbitals of S atom. Indicate angular and radial nodes in these orbitals. (4 points) b) Calculate Zeff experienced by and electron in 4p AO's in a arsenic atom. Use Slater rules that were discussed in lecture. (3 points)arrow_forward
- General Chemistry - Standalone book (MindTap Cour...ChemistryISBN:9781305580343Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; DarrellPublisher:Cengage LearningChemistry by OpenStax (2015-05-04)ChemistryISBN:9781938168390Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark BlaserPublisher:OpenStaxChemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage Learning
- Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage LearningChemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage LearningChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage Learning