A 30-kg iron block and a 40-kg copper block, both initially at 808C, are dropped into a large lake at 158C. Thermal equilibrium is

A 30-kg iron block and a 40-kg copper block, both initially at 808C, are dropped into a large lake at 158C. Thermal equilibrium is established after a while as a result of heat transfer between the blocks and the lake water. Determine the total entropy change for this process.

Answers

Explanation:

The given data is as follows.

    m_{1} = 30 kg,      m_{2} = 40 kg

   T_{1} = (80 - 15)^{o}C = 65^{o}C

   T_{2} = 65^{o}C

Now, we will calculate the heat energy as follows.

       Q_{1} = m_{1}C Delta T_{1}

                   = 30 times 0.45 times 65^{o}C

                   = 877.5 kJ

and,    Q_{2} = m_{2}C Delta T_{2}

                     = 40 times 0.386 times 65

                     = 1003.6 kJ

 (Delta S)_{lake} = frac{(Q_{1} + Q_{2})}{T_{o}}

               = frac{(877.5 + 1003.6) kJ}{288}

               = 6.531 kJ/K

 Delta S_{1} = m_{1}C_{1} ln frac{T_{2}}{T_{1}}

              = 30 times 0.45 ln frac{288}{353}

              = -2.747 kJ/K

 Delta S_{2} = m_{2}C_{2} ln frac{T_{2}}{T_{1}}

              = 40 times 0.386 times ln frac{288}{353}

              = -3.142 kJ/K

Now, we will calculate the total entropy as follows.

    Delta S_{total} = Delta S_{lake} + Delta S_{body}

                = 6.531 kJ/K - 2.747 kJ/K - 3.142 kJ/K

                = 0.64 kJ/K

thus, we can conclude that the total entropy change for this process is 0.64 kJ/K.

answer: h+ its less than oh

explanation:

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