Question

(a) What is the best coefficient of performance for a refrigerator that cools an environment at −30.0ºC and has heat transfer to another environment at 45.0ºC ? (b) How much work in joules must be done for a heat transfer of 4186 kJ from the cold environment? (c) What is the cost of doing this if the work costs 10.0 cents per 3.60×10⁶ J (a kilowatt-hour)? (d) How many kJ of heat transfer occurs into the warm environment? (e) Discuss what type of refrigerator might operate between these temperatures.

Short answer

The refrigerator’s performance coefficient is\(3.24\). 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 a warm environment. A very efficient refrigerator must work here which could efficiently decrease the temperature to -30 ^oC.

Step-by-step solution

Introduction

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.

 Given parameters & formula for efficiency of heat engine and performance coefficient

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’s performance coefficient\({\beta _{hp}} = \frac{1}{\eta }\)

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

Also, refrigerator’s 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’s performance coefficient.

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

\(Q\)-heat transfer.

\(W\) -work done.

 Calculate efficiency and coefficient of performance of refrigerator

(a) Maximum efficiency can be calculated by calculating Carnot efficiency as

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

Heat pump’s 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’s performance coefficient is

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

 Calculate work done

(b) Work done is

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

 Calculate total cost of this work

(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

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

 Calculate heat transferred to hot environment

(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’s 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.