Question

A 0.5kg piece of metal (c = 600/kgK) at 300 degree celcius is dumped into a large pool of water at 20 degrees celcius. Assuming the change in temperature of water to be negligible, calculate the overall change in entropy for the system
85.5J/K
67.4J/K
122.3J/K
202.3J/K

Answer 1

Answer on Question #41555, Physics, Molecular Physics | Thermodynamics

Question.

A $0.5\mathrm{kg}$ piece of metal $(c = 600 / \mathrm{kgK})$ at 300 degree celcius is dumped into a large pool of water at 20 degrees celcius. Assuming the change in temperature of water to be negligible, calculate the overall change in entropy for the system:

85.5J/K

67.4J/K

122.3J/K

202.3J/K

$m = 0.5kg$

$c = 600\frac{J}{kg\cdot K}$

$T_{1} = 300{}^{\circ}\mathrm{C} = 573K$

$T_{2} = 20{}^{\circ}\mathrm{C} = 293K$

$dS = ?$

Solution.

By definition entropy is:

$$\Delta S = \int \frac {\delta Q}{T}$$

In our case,

$$\delta Q = c m d T$$

Therefore,

$$\Delta S = \int_ {T _ {2}} ^ {T _ {1}} \frac {\delta Q}{T} = \int_ {T _ {2}} ^ {T _ {1}} c m \frac {d T}{T} = c m \ln \frac {T _ {1}}{T _ {2}}$$

So,

$$\Delta S = c m \ln \frac {T _ {1}}{T _ {2}} = 0. 5 \cdot 6 0 0 \cdot \ln \frac {5 7 3}{2 9 3} = 3 0 0 \cdot \ln 1. 9 5 6 = 3 0 0 \cdot 0. 6 7 1 \approx 2 0 2 \frac {J}{K}$$

Answer.

$$\Delta S = c m \ln \frac {T _ {1}}{T _ {2}} = 2 0 2. 3 \frac {J}{K}$$