Question

A 200 g insulated aluminum cup at 16 ∘C is filled with 255 g of water at 100 ∘C

a) Determine the final temperature of the mixture.

b)  Determine the total change in entropy as a result of the mixing process (use ΔS=∫dQ/T).

Answer 1

Solution :-

1. Find final temperature

Assumeing no losses the cup and water will end up at the final temp T2.

Specific heat from link

C aluminium – 0.215

C water – 1.00

Q cup + Q Water = 0

M cup C cup ΔT – cup + M water c water ΔT water = 0

200g * 0.215 * ΔT cup + 200g * 1 * ΔT water = 0

1. ΔT cup + 255 ΔT water = 0

43( T2 – T1 cup) + 255 (T2 – T1Water ) = 0

43T2 – 43T1 cup + 255T2 –T1 water = 0

298T2 = 43T1cup + 255T1 water

T2 = ( 43T1 cup + 255 T2 water ) / 298

T2 = (43*16⁰C + 255 * 100⁰C ) / 298

T2 = 688 + 25500 / 298

T2 = 87.8⁰C

The final temperature is = T2 = 87.8⁰C

       b ) Determine total change in enyropy

1. 16⁰C - 273 + 16 = 289 K
2. 87.8⁰C = 87.8 + 273 360.8 K
3. 100⁰C = 100 + 373 = 293K

Using formula ΔS = ∫ Dq / T

• S = ΔS * Cup + ΔS water
• S= Q cup / T aver + Q water / Tave

(MCΔT)-cupT ave + (MCΔT)-water /T ave

( 200*0.215*(360.8 – 289 ) /*(360.8 + 289 )/ 2

( 2555*1*(360.8 – 373 ) /*(360.8 +373 )/ 2 /2

30616 / 324.6 + 3111 / 366.9

9.4319162 – 8.4791

= 0.952

= ΔS

total change in entropy - 0.952