Please explain each answer as understandable as possible , please for full point

Question

Please explain each answer as understandable as possible , please for full points.

For parts a, b, and c, indicate the number of each answer that is correct. A piece of aluminum initially at 20.0 degree C is placed in a 110.degree C oven. The piece absorbs 85 kJ of energy on warming and the volume of the piece increases because of thermal expansion. The specific heat of aluminum is 0.900J/g degree C and the density of aluminum at 20.0 degree C is 2.70 g'mL. What is the value for q for the piece of aluminum? +85 kJ -85 kJ Slightly more positive than 85 kJ (90 kJ, for example) Slightly less positive than 85 kJ (80 kJ: for example) Slightly more negative than -85 kJ (-90 kJ, for example) Slightly less negative than -85 kJ (-80 kJ, for example) Cannot be determined What is the value for the change in internal energy for the piece of aluminum? Briefly justify your choice(s). +85 kJ -85 kJ Slightly more positive than 85 kJ (90 kJ, for example) Slightly less positive than 85 kJ (80 kJ: for example) Slightly more negative than -85 kJ (-90 kJ, for example) Slightly less negative than -85 kJ (-80 kJ, for example) Cannot be determined What is the value for the change in enthalpy for the piece of aluminum? Briefly justify your choice(s). +85 kJ -85 kJ Slightly more positive than 85 kJ (90 kJ, for example) Slightly less positive than 85 kJ (80 kJ: for example) Slightly more negative than -85 kJ (-90 kJ, for example) Slightly less negative than -85 kJ (-80 kJ, for example) Cannot be determined What is the mass of the piece of aluminum?

Explanation / Answer

a) i) +85kJ

This is because the aluminium piece absorbed 85kJ energy in the oven.

b) iv) Slightly less positive than +85kJ

This is because the solid already has a negative internal energy due to binding of atoms in crystal lattices. Addition of 85kJ to it makes it slightly less than 85kJ

c) i) +85 kJ

This is because, Delta(H) = Delta(U) + PDelta(V)

In this case, no work is being done.

=> Delta H = Delta U = +85kJ

d) Q =m*S*Delta(t)

=> 85000 = m *0.9000*90 => m = 85000/81 gm = 1049.38 g

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