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$$
\begin{array}{l}{\text { How much of the } 100 \mathrm{kJ} \text { of thermal energy at } 800 \mathrm{K} \text { can be converted to useful work? }} \\ {\text { Assume the environment to be at } 25^{\circ} \mathrm{C} .}\end{array}
$$

$$
\eta_{th, rev}=1-\frac{T_0}{T_H}
$$

$$
=1-\frac{298}{800}=0.6275
$$

$$
W_{\text {max,out }}=W_{\text {rev,out }}=\eta_{\text {th,rev }}Q_{in}
$$

$$
=0.6275 * 100
$$

$$
=62.75 \mathrm{kJ}
$$

$$
\begin{array}{l}{\text { A heat engine that receives heat from a furnace and rejects waste heat to a river at } 20^{\circ} \mathrm{C}} \\ {\text { has a thermal efficiency of } 40 \text { percent. Determine the second-law efficiency of this power plant. }}\end{array}
$$

$$
\eta_{\mathrm{II}}=\frac{\eta_{th}}{\eta_{th \ rev}}
$$

    $$
\eta_{th, rev}=1-\frac{T_{0}}{T_{H}}=1-\frac{20+273}{1200+273}
$$

$$
=0.801
$$

$$
\eta_{II}=\frac{0.4}{0.801}=49.9 \%
$$

$$
* 100 \%
$$

$$
\begin{array}{l}{\text { A house that is losing heat at a rate of } 50,000 \mathrm{kJ} / \mathrm{h} \text { when the outside temperature drops to } 4^{\circ} \mathrm{C} \text { is }} \\ {\text { to be heated by electric r resistance heaters. If the house is to be maintained at } 25^{\circ} \mathrm{C} \text { at all times, }} \\ {\text { determine the reversible work input for this process and the irreversibility. }}\end{array}
$$

$$
\dot{w}_{i n}=\dot{Q}_{out}=\dot{Q}_{H}
$$

$$
=\frac{50000 kJ/h}{3600} \longrightarrow 13.89 k w
$$

$$
\text { COP }_{HP, rev}=\frac{1}{1-T_L/T_H}
$$

$$
=\frac{1}{1-\frac{4+273}{25+273}}=14.2
$$

$$
\dot{W}_{\mathrm{rev \ in}}=
$$$$
\frac{\dot{Q}_{H}}{cop_{H P rev}}=\frac{13.89kw}{14.2}=0.978kw
$$

$$
\dot{I}=\dot{W}_{u \  i n}-\dot{W}_{rev \  i n}=13.89-0.978=12.91kw
$$

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