3. For parts a, b, and c, indicate the number of each answer that is correct. A

Question

3. For parts a, b, and c, indicate the number of each answer that is correct. A piece of aluminum initially at 20.0°C is placed in a 110. 'C oven. The piece absorbs 85.0 kJ of energy on warming and the volume of the piece increases because of thermal expansion. The fct of aluminum is o density of aluminum at 20.o*C is 2.70 g/mL. pansion. The specific heat of aluminum is 0.900 J/g C and the What is the value for q for the piece of aluminum? i. +85.0 k ii. -85.0 k ili. Slightly more positive than +85.0 kJ (+90 kJ, for example) iv. Slightly less positive than +85.0 kJ (+80 kJ, for example) v. Slightly more negative than-85.0 kJ (-90 kJ, for example) vi. slightly less negative than-85.0 k(-80 ki, for example) vii. Cannot be determined a.

Explanation / Answer

The process is isobaric process, since done in open at constant pressure.where q = delta H = delta U +P.delta V

a)option (i). in any process q = energy absorbed /given

here the system(Al piece) absorbed 85kJ energy. Thus q = + 85kJ which includes the change in internal nergy and the volume expansion energy.

b) Oprion (iv). As the process is isobaric the heat energy absorbed is utilised to increase the internal energy as well as to do mechanical(expansion) work. That is the increase in internal energy will be lower than +85kJ.

c) option (i) Since in an isobaric process q= delta H =change in enthalpy

we have delta H = +85kJ

d) When only pressure-volume work is done , no change in internal energy, thus q=w= -Pdelta V.

Since PV = nRT we can take 1atm (delta V ) = 1mol. 0.08206Latm/mol K x (110-20) =7.3854 L.atm

= 7.3854 x 101.26 J = 747,84J

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