Calculate the heat needed to raise 25.0 g of water by 10.0\ ^C (c = 4.18\ J g^-1\,^C^-1). [2 points]

A reaction releases 2.0\ kJ when 0.020 mol reacts. Calculate H per mole. [2 points]

College Board AP 2023 (style)-style practice: Section II (long FRQ, part). A 50.0 g sample of a metal at 100.0\ ^C is added to 100.0 g of water at 24.0\ ^C in a calorimeter. The final temperature is 28.0\ ^C. The specific heat of water is 4.18\ J g^-1\,^C^-1. (a) Calculate the heat gained by the water. (b) Determine the heat lost by the metal. (c) Calculate the specific heat capacity of the metal. (d) Justify why the calorimeter is assumed to be insulated.

A 4-point quantitative FRQ on calorimetry. (a) Heat gained by water (1 point): q_water = mc T = (100.0)(4.18)(28.0 - 24.0) = (100.0)(4.18)(4.0) = 1672\ J ≈ 1.67 × 10^3\ J. (b) Heat lost by metal (1 point): q_metal = -q_water = -1.67 × 10^3\ J (the metal loses what the water gains). (c) Specific heat of metal (1 point): T_metal = 28.0 - 100.0 = -72.0\ ^C; c = qm T = -1672(50.0)(-72.0) = 0.46\ J g^-1\,^C^-1. (d) Justify (1 point): assuming no heat escapes lets us set q_metal = -q_water; an insulated calorimeter ensures the only heat exchange is between the metal and the water. Markers reward the heat into water, equating it to the heat from the metal, the metal's specific heat, and the insulation assumption.

United StatesChemistrySyllabus dot point

How do we calculate the heat exchanged using specific heat capacity, and how does calorimetry measure it?

Topic 6.4 Heat Capacity and Calorimetry: use the equation q equals mc delta T with specific heat capacity, and use calorimetry data to determine the heat of a process.

A focused answer to AP Chemistry Topic 6.4, covering specific heat capacity, the equation q equals mc delta T, calorimetry, and how to determine the heat and enthalpy of a process from temperature data, with full worked examples.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
Specific heat capacity and the heat equation
Calorimetry
From heat to enthalpy per mole
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What this topic is asking

The College Board (Topic 6.4) wants you to use the equation $q = mc\Delta T$ with specific heat capacity to calculate the heat exchanged, and to use calorimetry data to determine the heat (and hence enthalpy) of a process. This is the quantitative core of the thermodynamics unit: turning a measured temperature change into a quantity of energy.

Specific heat capacity and the heat equation

The specific heat capacity is the energy needed to raise the temperature of one gram of a substance by one degree Celsius (or kelvin). A large $c$ (like water's $4.18\ \text{J g}^{-1}\,^\circ\text{C}^{-1}$ ) means the substance resists temperature change, absorbing a lot of heat for a small rise. This is why water is used as a coolant and why oceans moderate climate.

Calorimetry

A simple coffee-cup calorimeter is an insulated container of water in which a reaction occurs. You measure the temperature change of the water, compute $q_\text{water} = mc\Delta T$ , and infer the heat of the reaction as its negative. Dividing by the moles of the limiting reactant gives the molar enthalpy change $\Delta H$ . The insulation assumption (no heat lost to the surroundings) is what allows the simple energy balance.

From heat to enthalpy per mole

For a reaction at constant pressure, the heat measured is the enthalpy change for the amount of substance reacted. To report a molar enthalpy, divide the total heat by the number of moles:

\Delta H = \frac{q_\text{reaction}}{n}

The sign follows the convention: an exothermic reaction warms the water ( $q_\text{water} > 0$ ), so $q_\text{reaction} < 0$ and $\Delta H < 0$ . This links the laboratory measurement directly to the enthalpy of reaction defined in Topic 6.6.

Finding the molar enthalpy from calorimetry

Dissolving $0.0500$ mol of a salt in $200.0$ g of water raises the temperature from $25.0\ ^\circ\text{C}$ to $29.0\ ^\circ\text{C}$ . Find $\Delta H$ per mole ( $c_\text{water} = 4.18\ \text{J g}^{-1}\,^\circ\text{C}^{-1}$ ).

step 1 Heat gained by the water

$q_\text{water} = mc\Delta T = (200.0)(4.18)(29.0 - 25.0) = (200.0)(4.18)(4.0) = 3344\ \text{J}$ .

step 2 Heat of the dissolving process

$q_\text{reaction} = -q_\text{water} = -3344\ \text{J}$ (the process released heat, warming the water).

step 3 Convert to per mole

$\Delta H = \dfrac{q_\text{reaction}}{n} = \dfrac{-3344}{0.0500} = -66880\ \text{J mol}^{-1} = -66.9\ \text{kJ mol}^{-1}$ .

step 4 Interpret the sign

$\Delta H < 0$ , so the dissolution is exothermic, consistent with the water warming up.

Try this

Q1. Calculate the heat needed to raise $25.0$ g of water by $10.0\ ^\circ\text{C}$ ( $c = 4.18\ \text{J g}^{-1}\,^\circ\text{C}^{-1}$ ). [2 points]

Cue. $q = (25.0)(4.18)(10.0) = 1045\ \text{J} \approx 1.05 \times 10^{3}\ \text{J}$ .

Q2. A reaction releases $2.0\ \text{kJ}$ when $0.020$ mol reacts. Calculate $\Delta H$ per mole. [2 points]

Cue. $\Delta H = \dfrac{-2.0\ \text{kJ}}{0.020\ \text{mol}} = -100\ \text{kJ mol}^{-1}$ .

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)4 marksSection II (long FRQ, part). A

50.0

g sample of a metal at

100.0\ ^\circ\text{C}

is added to

100.0

g of water at

24.0\ ^\circ\text{C}

in a calorimeter. The final temperature is

28.0\ ^\circ\text{C}

. The specific heat of water is

4.18\ \text{J g}^{-1}\,^\circ\text{C}^{-1}

. (a) Calculate the heat gained by the water. (b) Determine the heat lost by the metal. (c) Calculate the specific heat capacity of the metal. (d) Justify why the calorimeter is assumed to be insulated.

Show worked answer →

A 4-point quantitative FRQ on calorimetry.

(a) Heat gained by water (1 point): $q_\text{water} = mc\Delta T = (100.0)(4.18)(28.0 - 24.0) = (100.0)(4.18)(4.0) = 1672\ \text{J} \approx 1.67 \times 10^{3}\ \text{J}$ .
(b) Heat lost by metal (1 point): $q_\text{metal} = -q_\text{water} = -1.67 \times 10^{3}\ \text{J}$ (the metal loses what the water gains).
(c) Specific heat of metal (1 point): $\Delta T_\text{metal} = 28.0 - 100.0 = -72.0\ ^\circ\text{C}$ ; $c = \dfrac{q}{m\Delta T} = \dfrac{-1672}{(50.0)(-72.0)} = 0.46\ \text{J g}^{-1}\,^\circ\text{C}^{-1}$ .
(d) Justify (1 point): assuming no heat escapes lets us set $q_\text{metal} = -q_\text{water}$ ; an insulated calorimeter ensures the only heat exchange is between the metal and the water.

Markers reward the heat into water, equating it to the heat from the metal, the metal's specific heat, and the insulation assumption.

AP 2021 (style)1 marksSection I (multiple choice). Equal masses of water and aluminum absorb the same quantity of heat. Water (

c = 4.18

) has a higher specific heat than aluminum (

c = 0.90\ \text{J g}^{-1}\,^\circ\text{C}^{-1}

). The temperature rise is (A) greater for water (B) greater for aluminum (C) equal for both (D) zero for both. Justify your choice.

Show worked answer →

A 1-point conceptual MCQ. The answer is (B).

From $q = mc\Delta T$ , with $q$ and $m$ fixed, $\Delta T = \dfrac{q}{mc}$ is inversely proportional to $c$ . Aluminum has the smaller specific heat, so it has the larger temperature rise. The trap is assuming the same heat gives the same temperature change regardless of $c$ .

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Sources & how we know this

AP Chemistry Course and Exam Description — College Board (2020)