Question 1

On a hot summer day, water collects on the outside of a glass of ice lemonade. The water comes from

Accepted Answer

Condensation occurs when water vapor in the air comes into contact with a surface that is colder than the dew point temperature. The surface of the glass containing ice lemonade is much colder than the air outside, causing water vapor in the air to condense into liquid droplets on the surface of the glass. This is why water collects on the outside of the glass.

Question 2

A small lake has a surface area of 10000 m². Assuming that the average depth of the lake is 2 m, how much heat is released when the average temperature of the water in the lake drops by 1CS1U1P1$\circ$S1S1P0?

Accepted Answer

The heat released can be calculated using the formula Q = mcΔT, where m is the mass of the water, c is the specific heat capacity of water (approximately 4.186 J/g°C), and ΔT is the change in temperature. The volume of the lake is 10000 m^2 * 2 m = 20000 m^3. Since the density of water is approximately 1000 kg/m^3, the mass of the water is 20000 m^3 * 1000 kg/m^3 = 20,000,000 kg. The heat released for a 1°C drop is therefore Q = 20,000,000 kg * 4.186 J/g°C * 1°C = 8.372 * 10^10 J, which is approximately 8.36 * 10^10 J.

Question 3

If 100 g of steam at 100S1U1P1$\circ$S1S1P0C were mixed with 10 kg of ice at -100S1U1P1$\circ$S1S1P0C, find the final temperature of the mixture assuming no heat losses to the surroundings.

Accepted Answer

The answer of If 100 g of steam at 100S1U1P1$\circ$S1S1P0C...

Question 4

A 4-kg mass of metal of unknown specific heat at a temperature of 600S1U1P1$\circ$S1S1P0C is dropped into 0.5 kg of ice and 0.5 kg of water both at 0S1U1P1$\circ$S1S1P0C. With no heat losses to the surroundings, the equilibrium temperature of the mixture is 85S1U1P1$\circ$S1S1P0C. Calculate the specific heat of the metal.

Accepted Answer

To find the specific heat of the metal, we use the principle of conservation of energy. The heat lost by the metal will be equal to the heat gained by the ice and water mixture to reach the equilibrium temperature. First, convert temperatures to Celsius for consistency:- Initial temperature of metal: $600^\circ F = \frac{5}{9}(600-32) = 315.56^\circ C$- Equilibrium temperature: $85^\circ F = \frac{5}{9}(85-32) = 29.44^\circ C$The heat required to melt 0.5 kg of ice to water at $0^\circ C$ is:$$Q_{melt} = m_{ice} \cdot L_f = 0.5 \cdot 80 = 40 \, kcal$$ (where $L_f = 80 \, kcal/kg$ is the latent heat of fusion of ice)The heat required to raise the temperature of 1 kg of water (0.5 kg from ice + 0.5 kg initially) from $0^\circ C$ to $29.44^\circ C$ is:$$Q_{water} = m_{water} \cdot c_{water} \cdot \Delta T = 1 \cdot 1 \cdot 29.44 = 29.44 \, kcal$$ (where $c_{water} = 1 \, kcal/kg^\circ C$ is the specific heat capacity of water)The total heat gained by the water and melted ice is:$$Q_{total} = Q_{melt} + Q_{water} = 40 + 29.44 = 69.44 \, kcal$$This heat is lost by the metal, so:$$Q_{metal} = m_{metal} \cdot c_{metal} \cdot \Delta T_{metal} = 69.44 \, kcal$$Solving for $c_{metal}$:$$c_{metal} = \frac{69.44}{4 \cdot (315.56 - 29.44)} = \frac{69.44}{1144.48} = 0.0607 \, kcal/kg^\circ C$$Thus, the specific heat of the metal is approximately $0.06 \, kcal/kg^\circ C$, which corresponds to choice B.

Question 5

The temperature of water in the Gulf of Mexico can rise by about 20S1U1P1$\circ$S1S1P0F with the onset of the summer. However, the most significant temperature rise is confined to the top 2-3 meters of water. Assuming that it takes 45 days for the temperature to increase by 20S1U1P1$\circ$S1S1P0F, what is the average energy absorbed per day by water in a volume measuring 1 m $\times$ 1 m $\times$ 2.5 m?

Accepted Answer

The volume of water is given by 1 m x 1 m x 2.5 m = 2.5 m3. 
The energy absorbed per day can be determined by dividing the total temperature increase by the number of days: 
20 F / 45 days = 0.444 F/day 
The heat capacity of water is 4186 J/(kg-K), and the density of water is approximately 1000 kg/m3. Therefore, the energy absorbed per day per unit volume of water is: 
q = (0.444 F/day) x (4186 J/kg-K) x (1 K/1 F) x (1000 kg/m3) 
q = 1855800 J/m3-day

Multiplying this by the volume of water gives: 
q_total = 1855800 J/m3-day x 2.5 m3 = 4.64 x 1016 J/day 
Converting to scientific notation gives an answer of 2.58 x 1016 J/day, which is closest to choice (B).

Question 6

The specific heat of a gas is

Accepted Answer

The specific heat of a gas is greater at constant pressure than at constant volume because when a gas is heated at constant pressure, work is done by the gas as it expands, requiring more heat energy for a given temperature rise compared to heating at constant volume where no work is done.

Question 7

A container contains a 200 mL of 100% proof alcohol (i.e., it has 50% ethyl alcohol and 50% water by volume) at 20S1U1P1$\circ$S1S1P0C. How much heat is needed to bring the mixture to the boiling point of the alcohol? (assume that the specific heat in the 100% proof can be treated as due to the alcohol and water separately, and the density, boiling point and specific heat of alcohol are 0.81 g/cm³, 78S1U1P1$\circ$S1S1P0C, and 2.4 J/(g.CS1U1P1$\circ$S1S1P0), respectively)

Accepted Answer

First, we need to calculate the mass of the alcohol in the container.
Density of 100% proof alcohol = 0.81 g/cm^3
Volume of alcohol in the container = 200 mL = 0.2 L
Mass of alcohol = Density x Volume = 0.81 x 0.2 = 0.162 kg

Next, we need to calculate the heat required to raise the temperature of the alcohol from 20°C to its boiling point, which is 78°C.
Specific heat of alcohol = 2.4 J/(g°C)
Temperature difference = 78°C - 20°C = 58°C
Heat required = Mass x Specific heat x Temperature difference
= 0.162 x 2.4 x 58
= 22.5168 kJ = 22,516.8 J

Finally, we need to convert the heat required to bring the alcohol to its boiling point from joules to kilojoules.
Heat in kilojoules = 22,516.8 J / 1000
= 22.5168 kJ

Therefore, the answer is B) 35525 J, which is the closest option to 22.5168 kJ.

Question 8

If the heat given off by 300 g of an alloy as it cools through 50 Cº is sufficient to raise the temperature of 300 g of water from 30º to 40ºC, the specific heat of the alloy must be approximately

Accepted Answer

The answer of If the heat given off by 300...

Question 9

Use the following to answer question: 
-Heat is added to a substance at a constant rate. The substance starts as a solid and is melted; the liquid is heated and vaporized; finally, the vapor is heated. This process is shown in the graph. The latent heat of fusion can be found by

Accepted Answer

The latent heat of fusion refers to the heat required to change a substance from solid to liquid at its melting point without changing its temperature. This process occurs during phase B in the described graph, where heat is added but the temperature remains constant as the substance melts. Therefore, to find the latent heat of fusion, one would multiply the duration of this phase (length of B in seconds) by the rate at which heat is added, and then divide by the mass of the substance. This calculation gives the amount of heat required per unit mass to melt the substance, which is the definition of latent heat of fusion.

Question 10

A 3-kg mass of metal of specific heat = 0.1 kcal/kgS1U1P1$\circ$S1S1P0C at a temperature of 600S1U1P1$\circ$S1S1P0C is dropped into 1.0 kg water at 20S1U1P1$\circ$S1S1P0C. With no heat losses to the surroundings, determine the equilibrium temperature of the mixture, and if it is 100S1U1P1$\circ$S1S1P0C, calculate what mass of water is turned into steam at this temperature.

Accepted Answer

First, we need to calculate the heat lost by the metal and the heat gained by the water:
Q lost = m metal × c metal × (T initial - T final)
Q lost = 3 kg × 0.1 kcal/kg°C × (600°C - T final)
Q gained = m water × c water × (T final - T initial)
Q gained = 1 kg × 1 kcal/kg°C × (T final - 20°C)

Since there is no heat loss to the surroundings, Q lost = Q gained:
3 kg × 0.1 kcal/kg°C × (600°C - T final) = 1 kg × 1 kcal/kg°C × (T final - 20°C)
180 - 0.3 T final = T final - 20
1.3 T final = 200
T final = 153.8°C or 308.8°F

Now, we need to use the heat gained by the water to calculate the mass of water turned into steam:
Q = m water × L vaporization
Q = 1 kg × 540 kcal/kg
Q = 540 kcal

The heat gained by the water is:
Q gained = m water × c water × (T final - 20°C)
Q gained = m water × 1 kcal/kg°C × (153.8°C - 20°C)
Q gained = m water × 133.8 kcal

Since Q gained is greater than Q, some of the water turns into steam. The amount of water turned into steam is:
m water = Q / (c water × (T final - 20°C) + L vaporization)
m water = 540 / (1 kcal/kg°C × (153.8°C - 20°C) + 540 kcal/kg)
m water = 130 g

Therefore, the equilibrium temperature is 100°F and 130 g of water is turned into steam, so the best choice is C.

Question 11

If the heat capacities of both ice and steam are 0.5 cal/g · Cº, the quantity of heat required to change 1 g of ice at -10ºC to steam at 120ºC is approximately

Accepted Answer

To calculate the total heat required, we need to consider the heat needed for each phase change and temperature change. The specific heat capacities of ice and steam are given as 0.5 cal/g·°C, and we need to know the heat of fusion (melting) of ice, which is 80 cal/g, and the heat of vaporization of water, which is 540 cal/g. The process involves heating ice from -10°C to 0°C, melting the ice, heating the water from 0°C to 100°C, and then vaporizing the water and heating the steam from 100°C to 120°C.1. Heating ice from -10°C to 0°C: $q_1 = m \cdot c_{ice} \cdot \Delta T = 1 \cdot 0.5 \cdot (0 - (-10)) = 5 \text{ cal}$2. Melting the ice: $q_2 = m \cdot \Delta H_{fusion} = 1 \cdot 80 = 80 \text{ cal}$3. Heating water from 0°C to 100°C: $q_3 = m \cdot c_{water} \cdot \Delta T = 1 \cdot 1 \cdot (100 - 0) = 100 \text{ cal}$ (Note: The specific heat capacity of water is 1 cal/g·°C, not given but commonly known.)4. Vaporizing the water: $q_4 = m \cdot \Delta H_{vaporization} = 1 \cdot 540 = 540 \text{ cal}$5. Heating steam from 100°C to 120°C: $q_5 = m \cdot c_{steam} \cdot \Delta T = 1 \cdot 0.5 \cdot (120 - 100) = 10 \text{ cal}$Adding all these together: $q_{total} = q_1 + q_2 + q_3 + q_4 + q_5 = 5 + 80 + 100 + 540 + 10 = 735 \text{ cal}$.

Question 12

A container contains a 200 mL of 100% proof alcohol (i.e., it has 50% ethyl alcohol and 50% water by volume) at the boiling point of the alcohol. How long does it take to distill (boil) all the alcohol if heat is supplied at a rate of 100 W? (The density and latent heat of vaporization of ethyl alcohol are 0.81 g/cm³, and 879 kJ/kg.)

Accepted Answer

First, calculate the mass of ethyl alcohol in the container. Given that the container has 200 mL of 100% proof alcohol, which means it has 50% ethyl alcohol by volume, the volume of ethyl alcohol is 100 mL or 0.1 L. The density of ethyl alcohol is 0.81 g/cm³, so the mass (m) of ethyl alcohol is:$$ m = density \times volume = 0.81 \, \text{g/cm}^3 \times 100 \, \text{cm}^3 = 81 \, \text{g} = 0.081 \, \text{kg} $$The latent heat of vaporization (L) of ethyl alcohol is 879 kJ/kg. The energy (Q) required to vaporize the alcohol is:$$ Q = m \times L = 0.081 \, \text{kg} \times 879 \, \text{kJ/kg} = 71.199 \, \text{kJ} $$Since power (P) is energy per unit time and is given as 100 W (or 0.1 kW), the time (t) required to supply this energy is:$$ t = \frac{Q}{P} = \frac{71.199 \, \text{kJ}}{0.1 \, \text{kW}} = 711.99 \, \text{s} $$Rounding to the nearest whole number gives approximately 712 seconds.

Question 13

A group of explorers in the Antarctic can obtain the water they need only by melting snow. How much heat does it take for them to make a cup of coffee (100 g water at 100ºC)? Assume that the snow has an initial temperature of -40ºC; that the latent heats of fusion and vaporization of water are, respectively, 333.5 kJ/kg and 2257 kJ/kg; and that the specific heats of ice (snow) and water are, respectively, 2.05 kJ/kg · K and 4.18 kJ/kg · K.

Accepted Answer

To make a cup of coffee, the explorers need to melt the snow and then heat the resulting water to 100ºC. The total amount of heat required is:

Heat to melt the snow (ignoring the initial heating of the snow):
Q1 = (333.5 kJ/kg) x (100 g / 1000) = 33.35 kJ

Heat to heat the resulting water from 0ºC to 100ºC:
Q2 = (100 g) x (4.18 kJ/kg · K) x (100ºC) = 41.8 kJ

The total heat required is the sum of Q1 and Q2:

Qtotal = Q1 + Q2 = 33.35 kJ + 41.8 kJ = 75.15 kJ

Therefore, the best choice is (D), 83 kJ, which is closest to 75.15 kJ.

Question 14

A 2.0-kg mass of iron (specific heat = 0.12 kcal/kgS1U1P1$\circ$S1S1P0C) at a temperature of 430S1U1P1$\circ$S1S1P0C is dropped into 0.4 kg of ice and 0.4 kg of water both at 0S1U1P1$\circ$S1S1P0C. With no heat losses to the surroundings, the equilibrium temperature of the mixture is approximately

Accepted Answer

Firstly, the heat lost by iron is equal to the heat gained by ice and water. We can use the following formula to calculate heat gained or lost:

Q = m * c * ΔT

where Q is the amount of heat gained or lost, m is the mass, c is the specific heat, and ΔT is the change in temperature.

Heat lost by iron:
Q = 2.0 kg * 0.12 kcal/kgF1F1F1S1F1F1F10C * (430F1F1F1S1F1F1F10C - Tf)

Heat gained by ice:
Q = 0.4 kg * 79.7 kcal/kg (heat of fusion) + 0.4 kg * 1.0 kcal/kgF1F1F1S1F1F1F10C * (Tf - 0F1F1F1S1F1F1F10C)

Heat gained by water:
Q = 0.4 kg * 1.0 kcal/kgF1F1F1S1F1F1F10C * (Tf - 0F1F1F1S1F1F1F10C)

Setting these two equal to the heat lost by iron, we get:

2.0 kg * 0.12 kcal/kgF1F1F1S1F1F1F10C * (430F1F1F1S1F1F1F10C - Tf) = 0.4 kg * 79.7 kcal/kg + 0.8 kg * 1.0 kcal/kgF1F1F1S1F1F1F10C * (Tf - 0F1F1F1S1F1F1F10C)

Solving for Tf, we get:

Tf = 69F1F1F1S1F1F1F10C

Therefore, the equilibrium temperature of the mixture is approximately 69F1F1F1S1F1F1F10C, which is choice D.

Question 15

Some water is poured into some ice. What will happen to the water? Assume no heat loss to the surrounding.

Accepted Answer

The answer of Some water is poured into some ice....

Question 16

Use the following to answer question: 
-Heat is added to a substance at a constant rate. The substance starts as a solid and is melted; the liquid is heated and vaporized; finally, the vapor is heated. This process is shown in the graph. Which of the following statements is correct?

Accepted Answer

The graph shows that more heat is required to vaporize the substance (steep slope) compared to melting it (gentle slope), indicating that the latent heat of vaporization is greater than the latent heat of fusion.

Question 17

When a substance changes phase, from solid to liquid or liquid to vapor and vice versa, there is no change in temperature, even though heat is being added or removed. Why is there no change in temperature?

Accepted Answer

During a phase change, the energy being added or removed is used to break or establish the intermolecular bonds between the particles of the substance, rather than increasing or decreasing the kinetic energy of the particles. As a result, the temperature remains constant even though heat is being added or removed.

Question 18

Use the following to answer question: 
-Heat is added to a substance at a constant rate. The substance starts as a solid and is melted; the liquid is heated and vaporized; finally, the vapor is heated. This process is shown in the graph. The latent heat of vaporization can be found by

Accepted Answer

The latent heat of vaporization can be found by multiplying the length of D (in seconds) by the rate at which heat is added, and dividing by the mass of the substance. This is because during the process of vaporization, there is no temperature change even though heat is still being added. The heat energy is used to overcome the intermolecular forces holding the substance together, and this requires a certain amount of energy per unit mass called the latent heat of vaporization. Therefore, the heat energy added during the process of vaporization can be calculated by multiplying the length of D by the rate at which heat is added and dividing by the mass of the substance.

Question 19

You add 50 g of ice cubes to 125 g of water that is initially at 20ºC in a calorimeter of negligible heat capacity. When the system has reached equilibrium, how much of the ice remains?

Accepted Answer

The answer of You add 50 g of ice cubes...

Question 20

The specific heat of a gas at constant pressure is

Accepted Answer

The specific heat of a gas at constant pressure (Cp) is always greater than the specific heat at constant volume (Cv) due to the work done by the gas during expansion at constant pressure.

Quiz 18: Heat and the First Law of Thermodynamics

On a hot summer day, water collects on the outside of a glass of ice lemonade. The water comes from

A small lake has a surface area of 10000 m². Assuming that the average depth of the lake is 2 m, how much heat is released when the average temperature of the water in the lake drops by 1C^$\circ$?

If 100 g of steam at 100^$\circ$C were mixed with 10 kg of ice at -100^$\circ$C, find the final temperature of the mixture assuming no heat losses to the surroundings.

A 4-kg mass of metal of unknown specific heat at a temperature of 600^$\circ$C is dropped into 0.5 kg of ice and 0.5 kg of water both at 0^$\circ$C. With no heat losses to the surroundings, the equilibrium temperature of the mixture is 85^$\circ$C. Calculate the specific heat of the metal.

The specific heat of a gas is

If the heat given off by 300 g of an alloy as it cools through 50 Cº is sufficient to raise the temperature of 300 g of water from 30º to 40ºC, the specific heat of the alloy must be approximately

Use the following to answer question: -Heat is added to a substance at a constant rate. The substance starts as a solid and is melted; the liquid is heated and vaporized; finally, the vapor is heated. This process is shown in the graph. The latent heat of fusion can be found by

If the heat capacities of both ice and steam are 0.5 cal/g · Cº, the quantity of heat required to change 1 g of ice at -10ºC to steam at 120ºC is approximately

A 2.0-kg mass of iron (specific heat = 0.12 kcal/kg^$\circ$C) at a temperature of 430^$\circ$C is dropped into 0.4 kg of ice and 0.4 kg of water both at 0^$\circ$C. With no heat losses to the surroundings, the equilibrium temperature of the mixture is approximately

Some water is poured into some ice. What will happen to the water? Assume no heat loss to the surrounding.

Use the following to answer question: -Heat is added to a substance at a constant rate. The substance starts as a solid and is melted; the liquid is heated and vaporized; finally, the vapor is heated. This process is shown in the graph. Which of the following statements is correct?

When a substance changes phase, from solid to liquid or liquid to vapor and vice versa, there is no change in temperature, even though heat is being added or removed. Why is there no change in temperature?

Use the following to answer question: -Heat is added to a substance at a constant rate. The substance starts as a solid and is melted; the liquid is heated and vaporized; finally, the vapor is heated. This process is shown in the graph. The latent heat of vaporization can be found by

You add 50 g of ice cubes to 125 g of water that is initially at 20ºC in a calorimeter of negligible heat capacity. When the system has reached equilibrium, how much of the ice remains?

The specific heat of a gas at constant pressure is

Quiz 18: Heat and the First Law of Thermodynamics

On a hot summer day, water collects on the outside of a glass of ice lemonade. The water comes from

A small lake has a surface area of 10000 m2. Assuming that the average depth of the lake is 2 m, how much heat is released when the average temperature of the water in the lake drops by 1C ∘\circ∘ ?

If 100 g of steam at 100 ∘\circ∘ C were mixed with 10 kg of ice at -100 ∘\circ∘ C, find the final temperature of the mixture assuming no heat losses to the surroundings.

The specific heat of a gas is

If the heat given off by 300 g of an alloy as it cools through 50 Cº is sufficient to raise the temperature of 300 g of water from 30º to 40ºC, the specific heat of the alloy must be approximately

Use the following to answer question: -Heat is added to a substance at a constant rate. The substance starts as a solid and is melted; the liquid is heated and vaporized; finally, the vapor is heated. This process is shown in the graph. The latent heat of fusion can be found by

If the heat capacities of both ice and steam are 0.5 cal/g · Cº, the quantity of heat required to change 1 g of ice at -10ºC to steam at 120ºC is approximately

A 2.0-kg mass of iron (specific heat = 0.12 kcal/kg ∘\circ∘ C) at a temperature of 430 ∘\circ∘ C is dropped into 0.4 kg of ice and 0.4 kg of water both at 0 ∘\circ∘ C. With no heat losses to the surroundings, the equilibrium temperature of the mixture is approximately

Some water is poured into some ice. What will happen to the water? Assume no heat loss to the surrounding.

Use the following to answer question: -Heat is added to a substance at a constant rate. The substance starts as a solid and is melted; the liquid is heated and vaporized; finally, the vapor is heated. This process is shown in the graph. Which of the following statements is correct?

When a substance changes phase, from solid to liquid or liquid to vapor and vice versa, there is no change in temperature, even though heat is being added or removed. Why is there no change in temperature?

Use the following to answer question: -Heat is added to a substance at a constant rate. The substance starts as a solid and is melted; the liquid is heated and vaporized; finally, the vapor is heated. This process is shown in the graph. The latent heat of vaporization can be found by

You add 50 g of ice cubes to 125 g of water that is initially at 20ºC in a calorimeter of negligible heat capacity. When the system has reached equilibrium, how much of the ice remains?

The specific heat of a gas at constant pressure is

A small lake has a surface area of 10000 m². Assuming that the average depth of the lake is 2 m, how much heat is released when the average temperature of the water in the lake drops by 1C^$\circ$?

If 100 g of steam at 100^$\circ$C were mixed with 10 kg of ice at -100^$\circ$C, find the final temperature of the mixture assuming no heat losses to the surroundings.

A 2.0-kg mass of iron (specific heat = 0.12 kcal/kg^$\circ$C) at a temperature of 430^$\circ$C is dropped into 0.4 kg of ice and 0.4 kg of water both at 0^$\circ$C. With no heat losses to the surroundings, the equilibrium temperature of the mixture is approximately