Question

A block of aluminum with m = 0.5 kg, T = 20oC is dropped into a reservoir at a temperature of 90oC. Calculate (a) the change in stored energy (ΔE), (b) the amount of heat transfer (Q), (c) the change in entropy (ΔS), (d) the amount of entropy transfer by heat and (e) the entropy generation (Sgen,univ) in the system's universe during the heat transfer process.

Answer 1

Step-by-step explanation:

The given data is as follows.

m = 0.5 kg,
`T = 20^(o)C`,
`T_(2) = 90^(o)C`

It is known that specific heat of aluminium is 0.91 kJ/kg.

As we know that, dQ = dU + dw

where, dQ = heat transfer

dU = change in internal energy

dw = work transfer

For the given system, work transfer "w" is 0.

(a) Hence, change in stored energy will be calculated as follows.

Q =
`mC \Delta T`

=
`0.5 * 0.91 * (90 - 20)`

= 31.85 kJ

(b) The amount of heat transferred will be equal to change in stored energy.

So, dQ = Q = 31.85 kJ

(c) Change in entropy will be calculated as follows.

dS =
`mC ln (T_(2))/(T_(1))`

=
`0.5 * 0.91 * ln (90)/(20)`

= 0.684 kJ/K

(d) Entropy transfer by heat will be calculated as follows.

`\Delta S = (dQ)/(dT)`

=
`(31.85)/((20 + 273))`

= 0.1087 kJ/K

(e) Entropy change will be calculated as follows.

Entropy change = entropy transfer + entropy generation

`S_(2) - S_(1) = (dQ)/(T) + S^(o)_(gen)`

0.684 kJ/K = 0.187 +
`S^(o)_(gen)`

`S^(o)_(gen)` = 0.5752 kJ/K