Question

In a container, $200 \,g$ of aluminum (specific heat $900\, J / kg K$ ) at $100^{\circ} C$ is mixed with $50\, g$ 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
$\therefore 0.05 \times 180+4.18 \times 10^{3} \times 0.05(t-20)$
$=0.2 \times 900(100-t)$
$209+180 t=1800+4180-9$
$389 t =22171$
$t=56.99^{\circ} C$
$t=57^{\circ} C$
$S_{1}=\frac{m Q}{T}+m c \log \left(\frac{T_{2}}{T_{1}}\right)$
$=\frac{0.05 \times 180}{273}+0.05 \times 4.18 \times 10^{3} \log \left(\frac{57+273}{20+273}\right)$
$=\frac{9}{273}+209 \log \frac{330}{293}$
$=0.032+209 \times 0.0596$
$=0.032+12.4564$
$=12.4884$
$S_{2}=m c \log \left(\frac{T_{2}}{T_{1}}\right)$
$=0.20 \times 900 \log \left(\frac{57+273}{100+273}\right)$
$=180 \log \frac{330}{373}$
$=-180 \times(0.0532)=-9.576$
So, change entropy $=S_{1}+S_{2}$
$=12.4884-9.576$
$=2.91 \,J / K \cong+2.8\, J / K$