What is Phases, heat & energy on the Chemistry Regents?

Phases, heat & energy is a Energy & Phases topic on the Chemistry Regents. RegentsLab has 37 practice questions on it, each with a full worked solution.

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Phases, heat & energy: Chemistry Regents Practice

Practice Phases, heat & energy for the Chemistry Regents with real-style questions and full worked solutions. Miss one? Read the solution — the exam reuses the same ideas every year.

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Question 61 · 2 pts

How much heat energy is absorbed when $50.0$ g of liquid water is heated from $20.0^\circ\text{C}$ to $30.0^\circ\text{C}$? (specific heat of water $= 4.18$ J/g$\cdot$K)

(A) $6270$ J
(B) $16{,}700$ J
(C) $209$ J
(D) $2090$ J ✓

Worked solution

$\Delta T = 30.0 - 20.0 = 10.0\ \text{K}$ Use the temperature change, not the final temperature.

$q = mC\Delta T = (50.0)(4.18)(10.0)$

$q = 2090\ \text{J}$

Answer: $2090$ J (about $2.09 \times 10^3$ J)

When a substance stays in one phase, added heat changes only its temperature, and $q=mC\Delta T$ depends on the temperature change, not the final temperature.

Question 65 · 2 pts

Which of the following phase changes is exothermic (releases energy to the surroundings)?

(A) Sublimation (solid $\rightarrow$ gas)
(B) Condensation (gas $\rightarrow$ liquid) ✓
(C) Melting (solid $\rightarrow$ liquid)
(D) Vaporization (liquid $\rightarrow$ gas)

Worked solution

$\text{gas} \rightarrow \text{liquid}: \text{particles lose energy}$ Moving to a more ordered phase gives off energy.

$\text{energy released} \Rightarrow \text{exothermic}$

Answer: Condensation (gas $\rightarrow$ liquid)

Phase changes toward a more ordered state (condensing, freezing, depositing) release energy and are exothermic; the reverse changes absorb energy.

Question 7 · 2 pts

The heat of vaporization of water is 2260 J/g. How much heat is required to vaporize 10.0 grams of water already at its boiling point?

(A) 22600 J ✓
(B) 2260 J
(C) 226 J
(D) 226000 J

Worked solution

$q = m \times H_v$

$q = 10.0\ g \times 2260\ J/g$

$q = 22600\ J$

Answer: 22600 J

The heat needed for a phase change at constant temperature is found by multiplying the mass by the heat of vaporization. Multiplying 10.0 g by 2260 J/g gives 22,600 J. A common error is misplacing a decimal point in the multiplication.

Question 18 · 1 pt

A sample releases heat to its surroundings as it changes phase. This process is

(A) endothermic, such as boiling
(B) endothermic, such as melting
(C) exothermic, such as evaporation
(D) exothermic, such as freezing ✓

Worked solution

$\text{releases heat} \Rightarrow \text{exothermic}$ Energy leaves the sample.

$\text{freezing, condensing release heat}$

Answer: \text{exothermic (freezing)}

Releasing heat is exothermic; freezing and condensation give off energy, while melting and boiling absorb it.

Question 62 · 2 pts

How much heat is required to melt $25.0$ g of ice at $0^\circ\text{C}$ into liquid water at $0^\circ\text{C}$? (heat of fusion of water $= 334$ J/g)

(A) $8350$ J ✓
(B) $334$ J
(C) $104.5$ J
(D) $56{,}500$ J

Worked solution

$q = mH_f = (25.0)(334)$ Melting occurs at constant temperature, so use the fusion equation.

$q = 8350\ \text{J}$

Answer: $8350$ J (about $8.35 \times 10^3$ J)

Melting is a phase change at constant temperature, so the heat is found with $q=mH_f$, not with the specific-heat equation.

Question 63 · 2 pts

How much heat is released when $15.0$ g of water vapor condenses to liquid water at $100^\circ\text{C}$? (heat of vaporization of water $= 2260$ J/g)

(A) $33{,}900$ J ✓
(B) $6270$ J
(C) $5010$ J
(D) $2260$ J

Worked solution

$q = mH_v = (15.0)(2260)$ Condensation releases the same energy per gram that vaporization absorbs.

$q = 33{,}900\ \text{J}$

Answer: $33{,}900$ J (about $3.39 \times 10^4$ J)

Condensation releases the same quantity of energy that vaporization absorbs, and it is calculated with $q=mH_v$.

Question 64 · 2 pts

A solid is heated at a constant rate until it melts. During the flat (plateau) portion of the heating curve where the solid is melting, which statement is correct?

(A) The temperature stays constant while the kinetic energy of the particles increases.
(B) The temperature increases while the kinetic energy of the particles increases.
(C) The temperature stays constant while the potential energy of the particles increases. ✓
(D) The temperature increases while the potential energy of the particles stays constant.

Worked solution

$\text{Plateau (melting): } T = \text{constant}$ A flat region on a heating curve marks a phase change.

$T\ \text{constant} \Rightarrow \overline{KE} = \text{constant}$ Average kinetic energy depends only on temperature.

$\text{absorbed heat} \rightarrow \Delta PE\ (\text{particles pulled apart})$

Answer: The temperature stays constant while the potential energy of the particles increases.

During a phase change the temperature (and thus average kinetic energy) stays constant; the absorbed energy raises the potential energy as particles are separated.

Question 40 · 2 pts

How much heat is absorbed when 50. grams of liquid water is heated from 20.°C to 40.°C? (The specific heat capacity of liquid water is $4.18\ \text{J/g}\cdot^{\circ}\text{C}$.)

(A) 4180 J ✓
(B) 8360 J
(C) 209000 J
(D) 2090 J

Worked solution

$q = mc\Delta T$ Heat equation

$\Delta T = 40.^{\circ}C - 20.^{\circ}C = 20.^{\circ}C$

$q = (50.\ g)(4.18\ \text{J/g}\cdot^{\circ}C)(20.^{\circ}C) = 4180\ \text{J}$

Answer: 4180 J

Using q = mcΔT with the given mass, specific heat, and temperature change of 20.°C, the heat absorbed is 4180 J. A common error is using only half the temperature change or forgetting to multiply by the mass, leading to answers like 2090 J.

Question 57 · 4 pts · constructed response

The heat of fusion of water is 334 J/g. Calculate the total amount of heat energy required to melt 25.0 grams of ice at its melting point, 0°C, with no change in temperature.

Model answer: 8350 J of heat energy are required.

Worked solution

$q = m \times H_f$ Heat energy for a phase change at constant temperature

$q = 25.0\ \text{g} \times 334\ \text{J/g} = 8350\ \text{J}$

Answer: 8350 J of heat energy are required.

During melting, all added heat energy goes into changing the phase of the substance rather than its temperature, so the heat required equals the mass multiplied by the heat of fusion.

Question 73 · 4 pts · constructed response

Determine the amount of heat, in joules, required to raise the temperature of 50.0 grams of water from 20.0°C to 60.0°C. The specific heat capacity of water is 4.18 J/(g·°C).

Model answer: 8,360 joules of heat are required.

Worked solution

$q = mc\Delta T$ Heat equation for temperature change

$\Delta T = 60.0^{\circ}C - 20.0^{\circ}C = 40.0^{\circ}C$ Find the change in temperature

$q = (50.0\ g)(4.18\ J/g^{\circ}C)(40.0^{\circ}C) = 8360\ J$ Substitute and solve

Answer: 8,360 joules of heat are required.

The heat needed to change the temperature of a substance (without a phase change) is found using q = mcΔT, multiplying the mass, specific heat capacity, and change in temperature.

Question 81 · 4 pts · constructed response

A 50.0 gram sample of water is heated from 20.0°C to 45.0°C. Using a specific heat capacity of water of $4.18 \; J/(g \cdot ^\circ C)$, determine the total amount of heat, in joules, absorbed by the water.

Model answer: The water absorbs 5225 joules of heat.

Worked solution

$q = mc\Delta T$ Heat equation

$\Delta T = 45.0^\circ C - 20.0^\circ C = 25.0^\circ C$

$q = (50.0 \text{ g})(4.18 \; J/(g\cdot^\circ C))(25.0^\circ C) = 5225 \text{ J}$

Answer: The water absorbs 5225 joules of heat.

The heat absorbed by a substance during a temperature change (with no phase change) is found using q = mcΔT, multiplying the mass, specific heat capacity, and change in temperature together.

Question 8 · 2 pts

A pure solid substance is being heated at its melting point and is changing from a solid to a liquid. While this phase change is occurring, the temperature of the substance

(A) increases steadily
(B) remains constant ✓
(C) increases, then decreases
(D) decreases steadily

Worked solution

$\text{At the melting point, heat added overcomes intermolecular forces}$

$\text{This increases potential energy, not average kinetic energy}$

$\text{Since temperature depends on average kinetic energy, it stays constant during the phase change}$

Answer: remains constant

During a phase change at constant pressure, added heat changes potential energy (breaking intermolecular attractions) rather than kinetic energy, so temperature remains constant until the phase change is complete.

Question 33 · 2 pts

What is the total amount of heat energy required to vaporize 10.0 grams of water at its boiling point? [The heat of vaporization of water is 2260 J/g.]

(A) 2260 J
(B) 33 400 J
(C) 22 600 J ✓
(D) 226 J

Worked solution

$q = m \times H_v$ Heat needed for a phase change at constant temperature

$q = 10.0\text{ g} \times 2260\text{ J/g} = 22\,600\text{ J}$

Answer: 22 600 J

During vaporization, all added heat goes into changing the phase rather than raising temperature, so the heat required equals mass times the heat of vaporization. Multiplying 10.0 g by 2260 J/g gives 22,600 J.

Question 44 · 2 pts

The heat of fusion of water is 334 J/g. How much heat is absorbed when 10. grams of ice at 0°C completely melts to liquid water at 0°C?

(A) 3340 J ✓
(B) 334 J
(C) 33400 J
(D) 33.4 J

Worked solution

$q = m H_f$

$q = (10. \text{ g})(334 \text{ J/g}) = 3340 \text{ J}$

Answer: 3340 J

During melting at a constant temperature, the heat absorbed equals mass times the heat of fusion (q = mHf). Multiplying 10. grams by 334 J/g gives 3340 J. This heat changes phase only, it does not change the temperature, since the ice remains at 0°C throughout melting.

Question 54 · 4 pts · constructed response

The heat of fusion of water is 334 J/g. Calculate the total number of joules of heat required to melt 25.0 grams of ice at 0 degrees Celsius.

Model answer: 8350 J (8.35 x 10^3 J) of heat are required.

Worked solution

$q = m \times H_f$

$q = 25.0\ g \times 334\ \frac{J}{g} = 8350\ J$

Answer: 8350 J (8.35 x 10^3 J) of heat are required.

During melting, all added heat goes into overcoming intermolecular attractions (phase change) rather than raising temperature, so q equals mass times the heat of fusion, given directly by Table reference data.

Question 64 · 4 pts · constructed response

Determine the total number of joules of heat needed to raise the temperature of 25.0 grams of water from 20.°C to 80.°C. The specific heat capacity of water is 4.18 J/(g·°C).

Model answer: 6270 J

Worked solution

$q = mC\Delta T$ Heat energy equation

$\Delta T = 80.^\circ C - 20.^\circ C = 60.^\circ C$ Find the temperature change

$q = (25.0\ \text{g})(4.18\ \text{J/g}^\circ\text{C})(60.^\circ C)$ Substitute values

$q = 6270\ \text{J}$ Multiply to find total heat

Answer: 6270 J

The heat absorbed by a substance undergoing a temperature change (no phase change) is found using q = mCΔT, where m is mass, C is specific heat capacity, and ΔT is the change in temperature.

Question 81 · 4 pts · constructed response

Calculate the amount of heat, in joules, needed to raise the temperature of 50.0 grams of liquid water from 20.0°C to 80.0°C. The specific heat capacity of water is $4.18\ J/(g \cdot ^{\circ}C)$.

Model answer: 12,540 J (1.25 x 10^4 J)

Worked solution

$\Delta T = 80.0^{\circ}C - 20.0^{\circ}C = 60.0^{\circ}C$ Find the temperature change

$q = mc\Delta T = (50.0\ g)(4.18\ J/(g\cdot ^{\circ}C))(60.0^{\circ}C)$ Apply the heat equation

$q = 12{,}540\ J$

Answer: 12,540 J (1.25 x 10^4 J)

The heat absorbed or released when a substance changes temperature without changing phase is calculated using q = mcΔT, where m is mass, c is specific heat capacity, and ΔT is the change in temperature.

Question 8 · 2 pts

How much heat is released when 10.0 grams of liquid water at 0°C freezes completely to ice at 0°C? (The heat of fusion of water is 334 J/g.)

(A) 334 J
(B) 33,400 J
(C) 3340 J ✓
(D) 33.4 J

Worked solution

$q = m \times H_f$

$q = 10.0 \text{ g} \times 334 \text{ J/g} = 3340 \text{ J}$

Answer: 3340 J

During freezing, heat equal to the heat of fusion is released for each gram of substance that changes phase, with no temperature change. Multiplying the mass (10.0 g) by the heat of fusion (334 J/g) gives 3340 J released.

Question 44 · 2 pts

How much heat, in joules, is required to raise the temperature of 50. grams of water from 20.°C to 40.°C? (specific heat of water = 4.18 J/(g·°C))

(A) 2090 J
(B) 8360 J
(C) 209 J
(D) 4180 J ✓

Worked solution

$q = mc\Delta T$

$\Delta T = 40.°C - 20.°C = 20.°C$

$q = (50.\ \text{g})(4.18\ \text{J/(g·°C)})(20.°C) = 4180\ \text{J}$

Answer: 4180 J

The heat required equals mass times specific heat times the temperature change. Substituting 50. g, 4.18 J/(g·°C), and a 20.°C change gives 4180 J.

Question 77 · 4 pts · constructed response

A 50.0-gram sample of liquid water is heated from 20.0°C to 45.0°C. Using the specific heat capacity of water, $4.18\ J/(g\cdot^{\circ}C)$, calculate the total amount of heat absorbed by the water.

Model answer: The water absorbs 5225 J (about 5.23 x 10^3 J) of heat.

Worked solution

$q = mc\Delta T$

$\Delta T = 45.0^{\circ}C - 20.0^{\circ}C = 25.0^{\circ}C$

$q = (50.0\ g)(4.18\ J/g\cdot^{\circ}C)(25.0^{\circ}C)$

$q = 5225\ J \approx 5.23\times 10^{3}\ J$

Answer: The water absorbs 5225 J (about 5.23 x 10^3 J) of heat.

The heat absorbed by a substance during a temperature change (no phase change occurring) is found using q = mcΔT, where m is mass, c is specific heat capacity, and ΔT is the change in temperature.

Question 19 · 2 pts

Using $q = mH_f$, how much heat energy is released when 10.0 grams of water freezes? The heat of fusion of water is 334 J/g.

(A) 3340 J ✓
(B) 33,400 J
(C) 33.4 J
(D) 334 J

Worked solution

$q = m \times H_f$

$q = 10.0\ \text{g} \times 334\ \text{J/g} = 3340\ \text{J}$

Answer: 3340 J

The heat released during freezing equals the mass of the substance multiplied by the heat of fusion: 10.0 g x 334 J/g = 3340 J.

Question 34 · 2 pts

The heat of fusion of water is 334 J/g. How many joules of heat are required to melt 10.0 grams of ice at its melting point?

(A) 3340 J ✓
(B) 33.4 J
(C) 334 J
(D) 33400 J

Worked solution

$q = m \times H_f$ Heat needed for a phase change at constant temperature

$q = 10.0 \text{ g} \times 334 \text{ J/g} = 3340 \text{ J}$

Answer: 3340 J

During melting, all added heat goes into breaking intermolecular forces rather than raising temperature, so q = m x Hf. Multiplying the mass (10.0 g) by the heat of fusion (334 J/g) gives 3340 J.

Question 48 · 2 pts

A pure solid substance is being heated at a constant rate. While the substance is melting at its melting point, the temperature of the substance

(A) increases steadily
(B) increases, then decreases
(C) remains constant ✓
(D) decreases steadily

Worked solution

$\text{During a phase change, added heat increases potential energy, not average kinetic energy}$

$\text{Temperature depends on average kinetic energy} \Rightarrow \text{temperature stays constant during melting}$

Answer: remains constant

While a substance is melting, the energy added is used to overcome intermolecular forces (increasing potential energy), not to speed up particle motion, so temperature remains constant until melting is complete.

Question 75 · 4 pts · constructed response

Determine the amount of heat, in joules, required to raise the temperature of 50.0 grams of liquid water from 20.0 degrees Celsius to 80.0 degrees Celsius. The specific heat capacity of water is 4.18 J/(g times degree Celsius).

Model answer: The heat required is 12,540 J (1.254 x 10^4 J).

Worked solution

$\Delta T = 80.0 - 20.0 = 60.0\ ^{\circ}\text{C}$

$q = mc\Delta T = (50.0\ \text{g})(4.18\ \text{J/g}^{\circ}\text{C})(60.0\ ^{\circ}\text{C})$

$q = 12{,}540\ \text{J}$

Answer: The heat required is 12,540 J (1.254 x 10^4 J).

The heat absorbed by a substance undergoing a temperature change (no phase change) is found using q = mcΔT, where m is mass, c is specific heat capacity, and ΔT is the change in temperature.

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