Q: A box weighing 24.07 kg is carried up six flights of stairs spanning 9.48 m. How much work (in millijoules) is done on the system, and what is the sign and units of the work?

The work is $W = mgh = 24.07 \times 9.8 \times 9.48 = 2238$ J, which is $2.238 \times 10^6$ mJ, and the sign is positive because energy is added to the system.

Q: A 1.588 cm$^2$ cross section of an animal's muscle is suspended on a 2.51 mm metal rod in the lab. The muscle is stretched by 1.13 cm. If the Hooke's Law constant is $2.0492$ N/cm, what is the work done on the muscle?

The work is $W = \frac{1}{2} k x^2 = \frac{1}{2} \times 2.0492 \times (1.13)^2 = 1.31$ J.

Q: A 189 g liquid of water is heated to 86.3$^\circ$C at 1 atm. Calculate the heat (in kilojoules) absorbed by the system and specify the sign and units.

The heat absorbed is $q = m c \Delta T = 189 \times 4.18 \times (86.3 - 25) = 47800$ J $= 47.8$ kJ, and the sign is positive.

Q: A hiker at high altitude boils water. If 1.216 L of water is boiled at 23$^\circ$C, how much energy (in kJ) is required for this process? (Heat of vaporization of water at 23$^\circ$C is $43.9$ kJ/mol)

Energy required $= \frac{1.216 \text{ L} \times 1000 \text{ g/L}}{18.02 \text{ g/mol}} \times 43.9 \text{ kJ/mol} = 2960$ kJ.

Q: If the hiker had to replace the heat of evaporating 1.216 L of water by metabolism of glucose and his temperature would not change, how many grams of glucose would he have to metabolize? (Heat of reaction at 23$^\circ$C is $-2816$ kJ/mol, molar mass of glucose is $180$ g/mol)

Grams of glucose $= \frac{2960 \text{ kJ}}{2816 \text{ kJ/mol}} \times 180 \text{ g/mol} = 189$ g.

Q: What is the work done by the system when 617 g of liquid water is evaporated to steam at 100$^\circ$C at 1 atm?

Work $= P \Delta V = 1 \text{ atm} \times (V_{\text{steam}} - V_{\text{liquid}})$, which is positive because the system expands.

Question 1

A box weighing 24.07 kg is carried up six flights of stairs spanning 9.48 m. How much work (in millijoules) is done on the system, and what is the sign and units of the work?

Accepted Answer

The work is $W = mgh = 24.07 	imes 9.8 	imes 9.48 = 2238$ J, which is $2.238 	imes 10^6$ mJ, and the sign is positive because energy is added to the system.

Question 2

A 1.588 cm$^2$ cross section of an animal's muscle is suspended on a 2.51 mm metal rod in the lab. The muscle is stretched by 1.13 cm. If the Hooke's Law constant is $2.0492$ N/cm, what is the work done on the muscle?

Accepted Answer

The work is $W = \frac{1}{2} k x^2 = \frac{1}{2} 	imes 2.0492 	imes (1.13)^2 = 1.31$ J.

Question 3

A 189 g liquid of water is heated to 86.3$^\circ$C at 1 atm. Calculate the heat (in kilojoules) absorbed by the system and specify the sign and units.

Accepted Answer

The heat absorbed is $q = m c \Delta T = 189 	imes 4.18 	imes (86.3 - 25) = 47800$ J $= 47.8$ kJ, and the sign is positive.

Question 4

A hiker at high altitude boils water. If 1.216 L of water is boiled at 23$^\circ$C, how much energy (in kJ) is required for this process? (Heat of vaporization of water at 23$^\circ$C is $43.9$ kJ/mol)

Accepted Answer

Energy required $= \frac{1.216 	ext{ L} 	imes 1000 	ext{ g/L}}{18.02 	ext{ g/mol}} 	imes 43.9 	ext{ kJ/mol} = 2960$ kJ.

Question 5

If the hiker had to replace the heat of evaporating 1.216 L of water by metabolism of glucose and his temperature would not change, how many grams of glucose would he have to metabolize? (Heat of reaction at 23$^\circ$C is $-2816$ kJ/mol, molar mass of glucose is $180$ g/mol)

Accepted Answer

Grams of glucose $= \frac{2960 	ext{ kJ}}{2816 	ext{ kJ/mol}} 	imes 180 	ext{ g/mol} = 189$ g.

Question 6

What would happen to the percentage increase in density of water at 4$^\circ$C if the pressure at the ocean depth of 3675 m increased? (Assume the density of seawater is $1.025$ kg/L and isothermal compressibility $\beta$ of liquid water at 4$^\circ$C is $4.95 	imes 10^{-10}$ Pa$^{-1}$)

Accepted Answer

The percentage increase in density would increase because higher pressure compresses the water more.

Question 7

A balloon filled with oxygen at 1 atm and 28.62 L is submerged under the ocean at 3675 m. Assuming ideal gas behavior, what would its volume be at 271.6 K?

Accepted Answer

Using $P_1 V_1 / T_1 = P_2 V_2 / T_2$, $V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}$, the volume would decrease due to increased pressure.

Question 8

What is the work done by the system when 617 g of liquid water is evaporated to steam at 100$^\circ$C at 1 atm?

Accepted Answer

Work $= P \Delta V = 1 	ext{ atm} 	imes (V_{	ext{steam}} - V_{	ext{liquid}})$, which is positive because the system expands.

Question 9

A forest at 25$^\circ$C and 1 atm is modeled to provide 9.8 kg of carbon per square meter per year. How much carbon is present in the atmosphere above each square meter of the Earth's surface if the atmosphere is 20% O$_2$ and 80% N$_2$?

Accepted Answer

The mass of carbon in the atmosphere above each square meter is much less than 9.8 kg, because most atmospheric mass is N$_2$ and O$_2$.

Question 10

If the rate of carbon utilization in a forest is doubled, how would it affect the amount of CO$_2$ in the atmosphere directly above the forest?

Accepted Answer

The amount of CO$_2$ would decrease because more carbon is being fixed by the forest.

Question 11

Biological organisms living at great depths in the ocean have evolved to equilibrate their internal pressures with the external hydrostatic pressure. What is the main reason for this adaptation?

Accepted Answer

To prevent being crushed by the immense external pressure.

Question 12

Estimate the hydrostatic pressure (in atm) in the ocean at depth 3675 m. Assume the density of seawater is $1.025$ kg/L.

Accepted Answer

Hydrostatic pressure $P = ho g h = 1.025 	imes 1000 	ext{ kg/m}^3 	imes 9.8 	ext{ m/s}^2 	imes 3675 	ext{ m} = 3.69 	imes 10^7$ Pa $\approx 364$ atm.

Question 13

A 129 g sample of liquid water is frozen to ice at 0$^\circ$C at 1 atm. What is the heat (in kJ) released by the system, and what is the sign and units?

Accepted Answer

Heat released $q = m \Delta H_{	ext{fus}} = 129 	ext{ g} 	imes \frac{1 	ext{ mol}}{18.02 	ext{ g}} 	imes 6.01 	ext{ kJ/mol} = 43.1$ kJ, and the sign is negative.

Question 14

Consider a balloon filled with oxygen at 1 atm and 28.62 L. If the balloon is submerged under the ocean at a depth of 3675 m at 271.6 K, what is the percentage increase in density of the O$_2$?

Accepted Answer

The density increases because the pressure increases, and percentage increase can be calculated using $\Delta \rho / \rho = \beta \Delta P$.

Question 15

Repeat the calculation for percentage increase in density of O$_2$ at depth, taking into account its deviation from ideal behavior. The van der Waals constants for O$_2$ are $a = 1.360$ L$^2$ atm/mol$^2$ and $b = 0.318$ L/mol. What effect does this have compared to the ideal gas calculation?

Accepted Answer

The density will be slightly higher than predicted by the ideal gas law due to intermolecular attractions.

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