Question

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1. We want to determine how much the room temperature increases when a kg of ice freezes.

Suppose you have a freezer that needs 1 J of energy for every 3 J of heat it removes.

How much thermal energy must be removed from 1 kg of water at room temperature?

How much electrical energy is used to freeze the ice?

What is the total energy, including waste heat, that is dumped into the kitchen?

If the kitchen contains 40 kg of air, how much will the temperature rise?

What is the efficiency of the freezer?

(The latent heat of water for freezing is 80 cal/gm°C, and the specific heat of air is about 0.2 cal/gm°C.  We know you could look it up; we thought we'd save you the time.)

2. When a .22-caliber rifle is fired, the expanding gas from the burning gunpowder creates a pressure behind the bullet. This pressure causes the force that pushes the bullet through the barrel. The barrel has a length of 0.61 m and an opening whose radius is 2.8  103 m. A bullet (mass  2.6  103 kg) has a speed of 370 m/s after passing through this barrel. Ignore friction and determine the average pressure of the expanding gas.

3. The volume of a gas is changed along the curved line between A and B in the drawing. Do not assume that the curved line is an isotherm or that the gas is ideal. (a) Find the magnitude of the work for the process, and (b) determine whether the work is positive or negative

4. The drawing refers to one mole of a monatomic ideal gas and shows a process that has four steps, two isobaric (A to B, C to D) and two isochoric (B to C, D to A). Complete the following table by calculating U, W, and Q (including the algebraic signs) for each of the four steps.

5.  Suppose you had 10 identical molecules enclosed by a box. At a given instant, one molecule has an energy of 100 Joules, and the others are all stationary.

(A) What is the average kinetic energy of the 10 molecules?

(B) Is this a situation of high or low entropy?

(C) What happens to the energy and entropy of the molecules in the box as time passes? Does your answer depend on the insulation of the box?

6. Which is more disordered: a 1000 word paper with 1 word misspelled, or a 2000 word paper with 2 mistakes? Compare the entropy of the two situations.

Answer 1

1)

Mass of water =1kg

Temperature of water \(\mathrm{T}=20^{\circ} \mathrm{C}\) (room temperature)

Amount of Heat required to be removed to convert water at \(20^{\circ} \mathrm{C}\) into ice

= Heat required to bring water to \(0^{\circ} \mathrm{C}+\) Heat required to convert it into ice at \(0^{\circ} \mathrm{C}\)

\(=\left[\mathrm{m}_{w}\left(4.168 \frac{\mathrm{J}}{\mathrm{gm}^{\circ} \mathrm{C}}\right)(\Delta \mathrm{T})\right]+\left[\mathrm{m}_{w}\left(334 \frac{\mathrm{J}}{\mathrm{gm}}\right)\right]\)

\(=\left[1000 \mathrm{gm}\left(4.168 \frac{\mathrm{J}}{\mathrm{gm}^{\circ} \mathrm{C}}\right)\left(20^{\circ} \mathrm{C}\right)\right]+\left[1000 \mathrm{gm}\left(334 \frac{\mathrm{J}}{\mathrm{gm}}\right)\right]\)

$$ =83360 \mathrm{~J}+334000 \mathrm{~J} $$

\(=417360 \mathrm{~J}\)

\(=41736 \mathrm{~kJ}\)

Therefore, the amount of heat required to be removed by the refrigerator is \(Q=417.36 \mathrm{~kJ}\)

2)

As the refrigerator needs \(1 \mathrm{~J}\) of electrical energy to remove \(3 \mathrm{~J}\) of heat

Electrical energy required to remove \(\mathrm{Q}=417360 \mathrm{~J}\) heat is

Electrical energy required \(=\frac{Q}{3}\)

$$ \begin{array}{l} =\frac{417360 \mathrm{~J}}{3} \\ =139120 \mathrm{~J} \\ =139.12 \mathrm{~kJ} \end{array} $$

Therefore, total electrical energy required is \(139.12 \mathrm{~kJ}\)

3)

Total Heat Dumped into the kitchen is due to the heat taken out from the water and the heat input to the refrigerator.

Heat \(_{\text {Room }}=\) Heat \(_{\text {Water } \rightarrow \text { ice }}+\) Heat \(_{\text {electical }}\)

Heat \(_{\text {Room }}=417360 \mathrm{~J}+139120 \mathrm{~J}\)

Heat \(_{\text {Room }}=556480 \mathrm{~J}\)

Heat \(_{\text {Room }}=556.48 \mathrm{~kJ}\)

Therefore, total heat energy dumped to the kitchen is \(556.48 \mathrm{~kJ}\)

4)

The total heat radiated to the room will increase the temperature of air inside the room

The total heat radiated is \(556480 \mathrm{~J}\)

The total heat radiated in calories \(=133001.912\)

The initial temperature of air is \(20^{\circ} \mathrm{C}\)

The mass of air inside room is \(40 \mathrm{~kg}\)

Specific heat capacity of air \(0.2 \frac{\mathrm{cal}}{\mathrm{gm}^{\circ} \mathrm{C}}\)

Therefore, the temperature rise is

\(\mathrm{Q}=\mathrm{m}_{\mathrm{a}} \mathrm{C}_{\mathrm{V}}\left(\mathrm{T}_{2}-20^{\circ} \mathrm{C}\right)\)

\(133001.912=(40,000 \mathrm{gms})\left(0.2 \frac{\mathrm{Cal}}{\mathrm{gm}^{0} \mathrm{C}}\right)\left(\mathrm{T}_{2}-20^{\circ} \mathrm{C}\right)\)

\(16.62^{\circ} \mathrm{C}=\left(\mathrm{T}_{2}-20^{\circ} \mathrm{C}\right)\)

\(\mathrm{T}_{2}=36.62^{\circ} \mathrm{C}\)

Therefore, the temperature will rise by \(16.62^{\circ} \mathrm{C}\), and the final temperature of the air will be \(\mathbf{T}_{2}=36.62^{\circ} \mathrm{C}\)

5)

$$ \begin{aligned} &\text { The efficiency of refrigerator is }\\ &\mathrm{COP}_{\mathrm{R}}=\frac{\text { Output }}{\text { Input Energy }}\\ &\mathrm{COP}_{\mathrm{R}}=\frac{417.36 \mathrm{~kJ}}{139.12 \mathrm{~kJ}}\\ &\mathrm{COP}_{R}=3 \end{aligned} $$