Question

An aluminum rod conducts 8.40 cal/s from a heat source maintained at 225°C to a large body of water at 22°C. Calculate the rate at which entropy increases in this process.

Short answer

The rate of increase of entropy is \(4.86 \times {10{ - 2}}\;\frac{{{\rm{J/K}}}}{{\rm{s}}}\).

Step-by-step solution

Concepts

The extracted heat from the heat source is coming at a constant rate.

The change in entropy is\(\Delta S = \frac{Q}{T}\)

Given data

The rate of heat transfer is \(Q = 8.40\;{\rm{cal/s}} = \left( {8.40 \times 4.186} \right)\;{\rm{J/s}}\).

The sourece temparature is \({T_1} = {225 \circ }{\rm{C}} = 498\;{\rm{K}}\).

The lower temperature is \({T_2} = {22 \circ }{\rm{C}} = 295\;{\rm{K}}\).

Calculation

The same amount of heat comes out from the heat source and is absorbed by the lower-temperature body.

Now the rate of change in entropy during the heat coming from the source is \(\Delta {S_1} = - \frac{Q}{{{T_1}}}\).

The rate of change in entropy during the absorption of heat is \(\Delta {S_2} = \frac{Q}{{{T_2}}}\).

Therefore the rate of change of net entropy is,

\(\begin{array}{c}\Delta S = \Delta {S_1} + \Delta {S_2}\\ = - \frac{Q}{{{T_1}}} + \frac{Q}{{{T_2}}}\\ = Q\left( {\frac{1}{{{T_2}}} - \frac{1}{{{T_1}}}} \right)\end{array}\)

Now substituting the values in the above equation, you get,

\(\begin{array}{c}\Delta S = Q\left( {\frac{1}{{{T_2}}} - \frac{1}{{{T_1}}}} \right)\\ = \left[ {\left( {8.40 \times 4.186} \right)\;{\rm{J/s}}} \right] \times \left( {\frac{1}{{295\;{\rm{K}}}} - \frac{1}{{498\;{\rm{K}}}}} \right)\\ = 4.86 \times {10{ - 2}}\;\frac{{{\rm{J/K}}}}{{\rm{s}}}\end{array}\)

Hence, the rate of increase of entropy is \(4.86 \times {10{ - 2}}\;\frac{{{\rm{J/K}}}}{{\rm{s}}}\).