Question 8bc90

Jun 26, 2015

The approximate net chenge in entropy for this system will be +1.1 J/K.

Explanation:

A change in entropy for a reversible process that takes place at constant temperature can be calculated using this equation

$\Delta S = \frac{Q}{T}$, where

$Q$ - heat absorbed/released during the process;
$T$ - the temperature at which this takes place.

In your case, temperature is not constant for neither the piece of aluminium, nor the water. However, if the change in temperature is small enough, you can use the average temperature to express the change in entropy for a process.

$\Delta S = \frac{Q}{\overline{T}}$, where

$\overline{T}$ - the average temperature.

So, the idea behind this problem is that the heat lost by the metal will be absorbed by the water.

$- {q}_{\text{metal" = q_"water}}$

The equation that establishes a relationship between heat gained/lost and change in temperature looks like this

$q = m \cdot c \cdot \Delta T$, where

$m$ - the mass of the substance;
$c$ - its specific heat;
$\Delta T$ - the change in temperature, defined as ${T}_{\text{final" - T_"initial}}$;

So, plug in your values into the above equation and solve for ${T}_{\text{final}}$.

$- {m}_{\text{metal" * c_"metal" * DeltaT_"metal" = m_"water" * c_"water" * DeltaT_"water}}$

$3.8 \cancel{\text{kg") * 900"J"/(cancel("kg") cancel(""^@cancel("C"))) * (30 - T_"final")cancel(""^@"C") = 1cancel("kg") * 4184"J"/(cancel("kg") cancel(""^@"C")) * (T_"final" - 20)cancel(""^@"C}}$

$3420 \cancel{\text{J") * (30-T_"final") = 4184cancel("J") * (T_"final} - 20}$

$7604 \cdot {T}_{\text{final" = 186280 => T_"final" = 186280/7604 = 24.5^@"C}}$

The heat lost by the metal will be

${q}_{\text{metal" = 3.8 * 900 * (24.5-30) = -"18,810 J}}$

The heat gained by the water will be

${q}_{\text{water" = -q_"metal" = +"18,810 J}}$

Since the change in temperature is not that significant, you can determine the average temperature for the metal and for water by

$\left\{\begin{matrix}{T}_{\text{initial metal" = (273.15 + 30)"K" = "303.15 K" \\ T_"final" = (273.15 + 24.5)"K" = "297.65 K}}\end{matrix}\right.$

${\overline{T}}_{\text{metal" = ((303.15 + 297.65)"K")/2 = "300.4 K}}$

and

$\left\{\begin{matrix}{T}_{\text{initial water" = (273.15 + 20)"K" = "293.15 K" \\ T_"final" = "297.65 K}}\end{matrix}\right.$

${\overline{T}}_{\text{water" = ((293.15 + 297.65)"K")/2 = "295.4 K}}$

So, the net change in entropy for the metal will be

$\Delta {S}_{\text{metal" = Q_"metal"/bar(T)_"metal" = (-"18,810 J")/("300.4 K") = -"62.6 J/K}}$

The net change in entropy for the water will be

$\Delta {S}_{\text{water" = Q_"water"/bar(T)_"water" = (+"18,810 J")/("295.4 K") = +"63.7 J/K}}$

The net change in entropy for the system will thus be

$\Delta {S}_{\text{sys" = DeltaS_"metal" + DeltaS_"water}}$

DeltaS_"sys" = -62.6 + 63.7 = color(green)(+"1.1 J/K")#

SIDE NOTE I'll leave the answer rounded to two sig figs.