91影视

We calculate the coefficient of performance of refrigerator using the formula for efficiency of Carnot engine as we know that maximum efficiency is for Carnot engine when the temperature of the cold and hot environment is given and coefficient of performance of heat pump is reciprocal of this efficiency. We further work done and cost of this work. We then find heat transferred to a warm environment.

The temperature of cold environment\[{T_c} = {\rm{ - 30}}{\;^{\rm{o}}}{\rm{C}} = {\rm{243}}\;{\rm{K}}\]

The temperature of hot environment \[{T_h} = 45{\;^{\rm{o}}}{\rm{C}} = 318\;{\rm{K}}\]

Efficiency of Carnot engine \[h = {\rm{1}} - \frac{{{T_c}}}{{{T_h}}}\]

Heat pump鈥檚 performance coefficient\({\beta _{hp}} = \frac{1}{\eta }\)

Refrigerator鈥檚 performance coefficient\({\beta _{ref}} = {\beta _{hp}} - 1\)

Also, refrigerator鈥檚 performance coefficient\({\beta _{ref}} = \frac{Q}{W}\)

Here,

\(\eta \)- efficiency of engine.

\({T_c}\)- temperature of cold reservoir.

\({T_h}\)- temperature of hot reservoir.

\({\beta _{hp}}\)- heat pump鈥檚 performance coefficient.

\({\beta _{ref}}\)- refrigerator鈥檚 performance coefficient.

\(Q\)-heat transfer.

\(W\) -work done.

\[\begin{array}{c}h = 1 - \frac{{{T_c}}}{{{T_h}}}\\ = {\rm{1 - }}\frac{{{\rm{243}}}}{{{\rm{318}}}}\\ = {\rm{0}}{\rm{.236}}\end{array}\]

Heat pump鈥檚 performance coefficient can be calculated as

\[\begin{array}{c}{b_{hp}} = 1/h \\ = {\rm{1/0}}{\rm{.236}}\\ = {\rm{4}}{\rm{.24}}\end{array}\]

Refrigerator鈥檚 performance coefficient is

\[\begin{array}{c}{b_{ref}} = {b_{hp}} - {\rm{1}}\\ = {\rm{4}}{\rm{.24 - 1}}\\ = {\rm{3}}{\rm{.24}}\end{array}\]

(b) Work done is

\[\begin{array}{c}W = Q/{b_{hp}}\\ = {\rm{4186/3}}{\rm{.24 kJ}}\\ = {\rm{1291}}{\rm{.98 kJ}}\end{array}\]

(C) Cost of\(3.6 \times {10^6}\;{\rm{J}}\) work is\(10\;{\rm{cents}}\)

Cost of\(1\,\;{\rm{J}}\) work\[ = \frac{{{\rm{10}}}}{{{\rm{3}}{\rm{.6 \times 1}}{{\rm{0}}^{\rm{6}}}}}{\rm{ cents}}\]

Cost of\(1291.98\;{\rm{kJ}}\) work

\[\begin{array}{c}{\rm{cost}} = \frac{{{\rm{10 \times 1}}{\rm{.292 \times 1}}{{\rm{0}}^{\rm{6}}}}}{{{\rm{3}}{\rm{.6 \times 1}}{{\rm{0}}^{\rm{6}}}}}{\rm{ cents}}\\ = {\rm{3}}{\rm{.59 cents}}\end{array}\]

(d) Heat transferred to warm environment

\[\begin{array}{c}{Q_h} = W + {Q_c}\\ = {\rm{(1291}}{\rm{.98 + 4186) kJ}}\\ = {\rm{5477}}{\rm{.98 kJ}}\end{array}\]

(e) A very efficient refrigerator must work here which could efficiently decrease temperature to\( - 30\;{\rm{^\circ C}}\).

Therefore, the refrigerator鈥檚 performance coefficient is\(4.61\). The work that is needed to be done is\(1291.98\;{\rm{kJ}}\). The cost of doing this is\(3.59\)cents and\(5477.98\;{\rm{kJ}}\) heat is transferred to warm environment.