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An electric water heater having a 200-L capacity heats water from 23 to 55C.23 \text { to } 55^{\circ} \mathrm{C}. Heat transfer from the outside of the water heater is negligible, and the states of the electrical heating element and the tank holding the water do not change significantly. Perform a full exergy accounting, in kJ, of the electricity supplied to the water heater. Model the water as incompressible with a specific heat c=4.18kJ/kgK.c=4.18 \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}. Let T0=23C.T_{0}=23^{\circ} \mathrm{C}.

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Answered 2 years ago
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Data:

  • V=200lV=200\mathrm{\,l} -Tank volume.
  • t1=23Ct_1=23\mathrm{\,C} -Inital temperature.
  • t2=55Ct_2=55\mathrm{\,C} -Final temperature.
  • c=4.18kJkgKc=4.18\mathrm{\,\dfrac{kJ}{kg\cdot K}} -Sepcific heat.
  • T0=296KT_0=296\mathrm{\,K} -Dead state temperature.

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