Question

In a container, $200g$ of aluminum (specific heat $900J/kgK$ ) at $100^{\circ }C$ is mixed with $50g$ of water at $20^{\circ }C$, with the mixture thermally isolated. Find the entropy change of the aluminumwater system.

• A. +2.8 J/K
• B. -22.1 J/K
• C. +24.9 J/K
• D. None of the above

Answer 1

Answer: (A)

Let the final temperature of the system be $t^{o}C$
Heat lost = Heat gain
$\therefore 0.05\times 180+4.18\times 10^3\times 0.05(t-20)$
$=0.2\times 900(100-t)$
$209+180t=1800+4180-9$
$389t=22171$
$t=56.99^{\circ }C$
$t=57^{\circ }C$
$S_1=\frac{mQ}{T}+mc\log \left(\frac{T_2}{T_1}\right)$
$=\frac{0.05\times 180}{273}+0.05\times 4.18\times 10^3\log \left(\frac{57+273}{20+273}\right)$
$=\frac{9}{273}+209\log \frac{330}{293}$
$=0.032+209\times 0.0596$
$=0.032+12.4564$
$=12.4884$
$S_2=mc\log \left(\frac{T_2}{T_1}\right)$
$=0.20\times 900\log \left(\frac{57+273}{100+273}\right)$
$=180\log \frac{330}{373}$
$=-180\times (0.0532)=-9.576$
So, change entropy $=S_1+S_2$
$=12.4884-9.576$
$=2.91J/K\cong +2.8J/K$