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Chapter 15: Q60P (page 412)

A 鈥淐arnot鈥� refrigerator (the reverse of a Carnot engine) absorbs heat from the freezer compartment at a temperature of -17掳C and exhausts it into the room at 25掳C.

(a) How much work would the refrigerator do to change 0.65 kg of water at 25掳C into ice at -17掳C.

(b) If the compressor output is 105 W and runs 25% of the time, how long will this take?

Short Answer

Expert verified

(a) The work done by the refrigerator is\(5.03 \times {104}\;{\rm{J}}\).

(b) The time taken to do the work is\(32\;\min \).

Step by step solution

01

Heat removed from the water

In this problem, there are three parts in which the heat rejected is evaluated. First, the cooling of water up to the freezing point; second, freezing the liquid water; and last, cooling the ice.

02

Given data

The temperature of the freezer is \({T_{\rm{1}}} = - 17\circ {\rm{C}}\).

The temperature of the room is \({T_{\rm{2}}} = 25\circ {\rm{C}}\).

The mass is \(m = 0.65\,{\rm{kg}}\).

The output power is \(P = 105\,{\rm{W}}\).

03

Evaluation of heat rejected from the water

The relation to find the heat rejected is given by:

\(\begin{array}{l}Q = mc\Delta T + mL + mc'\Delta T'\\Q = m\left( {c\Delta T + L + c'\Delta T'} \right)\end{array}\)

Here, \(c\)and \(c'\) are the specific heat for water and ice, \(\Delta T\) and \(\Delta T'\) are the change in temperature for water and ice, and Lis the latent heat of water.

Substitute the values in the above expression.

\(\begin{array}{l}Q = \left( {0.65\;{\rm{kg}}} \right)\left( {\left( {4186\;{\rm{J/kg}} \cdot \circ {\rm{C}}} \right)\left( {25\circ {\rm{C}}} \right) + \left( {3.33 \times {{10}5}\;{\rm{J/kg}}} \right) + \left( {2100\;{\rm{J/kg}} \cdot \circ {\rm{C}}} \right)\left( {17\circ {\rm{C}}} \right)} \right)\\Q = 3.07 \times {105}\;{\rm{J}}\end{array}\)

04

(a) Evaluation of the work done by the refrigerator

The relation of efficiency is given by:

\(\begin{array}{c}\eta = \frac{W}{{W + Q}}\\1 - \frac{{{T_{\rm{1}}}}}{{{T_{\rm{2}}}}} = \frac{W}{{W + Q}}\\W = Q\left( {\frac{{{T_2}}}{{{T_1}}} - 1} \right)\end{array}\)

Substitute the values in the above expression.

\(\begin{array}{l}W = \left( {3.07 \times {{10}5}\;{\rm{J}}} \right)\left( {\frac{{\left( {25\circ {\rm{C}} + 273} \right)\;{\rm{K}}}}{{\left( { - 17\circ {\rm{C}} + 273} \right)\;{\rm{K}}}} - 1} \right)\\W = 5.03 \times {104}\;{\rm{J}}\end{array}\)

Thus, the work done by the refrigerator is \(5.03 \times {104}\;{\rm{J}}\).

05

(b) Evaluation of the time consumed by using compressor power

The relation to find the time is given by:

\(t = \frac{W}{P}\)

Substitute the values in the above expression.

\(\begin{array}{l}t = \left( {\frac{{5.03 \times {{10}4}\;{\rm{J}}}}{{105\,{\rm{W}}\left( {0.25} \right)}}} \right)\\t = 1916.1\;{\rm{s}}\\t \approx 32\;\min \end{array}\)

Thus, the time taken to do the work is \(32\;\min \).

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Most popular questions from this chapter

An ideal gas undergoes an adiabatic expansion, a process in which no heat flows into or out of the gas. As a result,

(a) the temperature of the gas remains constant and the pressure decreases.

(b) both the temperature and pressure of the gas decrease.

(c) the temperature of the gas decreases and the pressure increases.

(d) both the temperature and volume of the gas increase.

(e) both the temperature and pressure of the gas increase

Question: In an isobaric compression of an ideal gas,

(a) no heat flows into the gas.

(b) the internal energy of the gas remains constant.

(c) no work is done on the gas.

(d) work is done on the gas.

(e) work is done by the gas.

(II) When\({\bf{5}}{\bf{.80 \times 1}}{{\bf{0}}{\bf{5}}}\;{\bf{J}}\)of heat is added to a gas enclosed in a cylinder fitted with a light frictionless piston maintained at atmospheric pressure, the volume is observed to increase from\({\bf{1}}{\bf{.9}}\;{{\bf{m}}{\bf{3}}}\)to\({\bf{4}}{\bf{.1}}\;{{\bf{m}}{\bf{3}}}\). Calculate

(a) the work done by the gas, and

(b) the change in internal energy of the gas.

(c) Graph this process on a PV diagram.

Question:(II) A certain power plant puts out 580 MW of electric power. Estimate the heat discharged per second, assuming that the plant has an efficiency of 32%.

Question: An ideal gas undergoes an isobaric compression and then an isovolumetric process that brings it back to its initial temperature. Had the gas undergone one isothermal process instead,

(a) the work done on the gas would be the same.

(b) the work done on the gas would be less.

(c) the work done on the gas would be greater.

(d) Need to know the temperature of the isothermal process.
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