Quiz5KEY

Quiz5KEY - Chem 6A 2011 (Sailor) Name: Student ID Number:...

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Chem 6A 2011 (Sailor) QUIZ #5 Name: VERSION A KEY Student ID Number: Section Number: Some useful constants and relationships: Specific heat capacities (in J/g . K): H 2 O (l) = 4.184; Al (s) = 0.900; Cu (s) = 0.387; Steel (s) = 0.45 101.325 J = 1 L . atm 1 atm = 760 Torr 1J = 1kg . m 2 /s 2 1 eV = 1.6022 x 10 -19 J R = Ideal gas constant: 0.08206 L . atm . mol -1 . K -1 = 8.31451 J . mol -1 . K -1 Avogadro constant: 6.022 x 10 23 mole -1 Planck's constant = h = 6.6261 x 10 -34 J . s c = speed of light: 3.00 x 10 8 m/s R H = 1.097 x 10 -2 nm -1 C 2 = second radiation constant = 1.44 x 10 -2 K . m q = C p m Δ T T λ max = 1 5 C 2 Emitted power (W) Surface area (m 2 ) = ( cons tan t ) T 4 e = mc 2 c = λν 1 = R H 1 n 1 2 1 n 2 2 Λ Ν Μ Ξ Π Ο E = h ν E = hc E (in Joules) = 2.18 × 10 18 Z 2 n 2 Λ Ν Μ Ξ Π Ο
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QUIZ #5 1. (a) Calculate the heat (q) when 22.0 g of solid water ice cools from -11.0 °C to -97.0 °C. Set up but do not solve; circle your answer. The following table of thermochemical data for H 2 O may be useful. (20 pts) this was problem 6.23 in the text Heat of fusion 6 kJ/mol Heat of vaporization 48 kJ/mol heat capacity of H 2 O (s) 2.087 J/g . K heat capacity of H 2 O (l) 4.184 J/g . K heat capacity of H 2 O (g) 2.080 J/g . K standard heat of formation of H 2 O (l) -241.8 kJ/mol q = C p m Δ T = 2.087 J g K × 22.0 g × ( 97.0 + 11) Numerically this works out to -3,949 J. 15 pts for correct setup. OK if they get the sign wrong. -2 pts if they use the wrong value of heat capacity (4.184 is for liquid water, 2.080 is for gas phase water) (b) Is this process exothermic or endothermic (circle one)? exothermic 5 pts all or nothing 2. Calculate the final temperature when a steel bolt, initially at 100.0 °C and weighing 50.0 g, is placed in 37 g of liquid water, initially at 11.0 °C. The relevant thermochemical data are given on the cover page of this quiz. (10 pts) this was a variation of problems 6.26 and 6.28 in the text heat lost by bolt = - (heat gained by water), T final is the same for both applying the expression q = C p m Δ T to both, C bolt m bolt Δ T = C water m water Δ T C bolt m bolt ( T final T bolt ) = C water m water ( T final T water ) multiplying through, C bolt m bolt T final C bolt m bolt T bolt = C water m water T final + C water m water T water collect terms, simplify, C bolt m bolt T final + C water m water T final = C water m water T water + C bolt m bolt T bolt T final ( C bolt m bolt + C water m water ) = C water m water T water + C bolt m bolt T bolt solve for T final T final = C water m water T water + C bolt m bolt T bolt C bolt m bolt + C water m water
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Quiz5KEY - Chem 6A 2011 (Sailor) Name: Student ID Number:...

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