AP Chemistry Quiz: Heat Capacity And Calorimetry
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Heat Capacity And CalorimetryQuestion 1 of 20

A 40.0 g40.0\,\text{g} sample of aluminum is heated, and its temperature increases from 25.0∘C25.0^\circ\text{C} to 50.0∘C50.0^\circ\text{C}. The specific heat capacity of aluminum is 0.900 J (g⋅∘C)−10.900\,\text{J}\,(\text{g}\cdot{}^\circ\text{C})^{-1}. What is the heat absorbed by the aluminum sample, qq?

+900 J
+360 J
−900 J
+90.0 J
+1.80×10^3 J
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AP Chemistry Quiz

AP Chemistry Quiz: Heat Capacity And Calorimetry

Practice Heat Capacity And Calorimetry in AP Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Heat Capacity And Calorimetry, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A 40.0 g40.0\,\text{g} sample of aluminum is heated, and its temperature increases from 25.0∘C25.0^\circ\text{C} to 50.0∘C50.0^\circ\text{C}. The specific heat capacity of aluminum is 0.900 J (g⋅∘C)−10.900\,\text{J}\,(\text{g}\cdot{}^\circ\text{C})^{-1}. What is the heat absorbed by the aluminum sample, qq?

  1. +900 J (correct answer)
  2. +360 J
  3. −900 J
  4. +90.0 J
  5. +1.80×10^3 J
Explanation: This problem tests the skill of heat capacity and calorimetry. For aluminum heating from 25.0°C to 50.0°C, we use q = mcΔT with m = 40.0 g, c = 0.900 J/(g·°C), and ΔT = 50.0°C - 25.0°C = 25.0°C. Calculating: q = (40.0 g)(0.900 J/(g·°C))(25.0°C) = 900 J. Since temperature increases, the aluminum absorbs heat, so q = +900 J. A common error is using only one temperature value (like 25.0°C) instead of calculating the temperature difference, which would give +360 J. Always calculate ΔT as the difference between final and initial temperatures before substituting into q = mcΔT.

Question 2

A reaction occurs in a coffee-cup calorimeter and causes the temperature of the solution to decrease from 26.0∘C26.0^\circ\text{C} to 23.0∘C23.0^\circ\text{C}. The total heat capacity of the solution is Csoln=500 J / ∘CC_{\text{soln}}=500\,\text{J}\,/\,^\circ\text{C}. What is the heat transferred for the solution, qsolnq_{\text{soln}}?

  1. +1.00×10^3 J
  2. −1.67×10^2 J
  3. −5.00×10^2 J
  4. +1.50×10^3 J
  5. −1.50×10^3 J (correct answer)
Explanation: This problem tests the skill of heat capacity and calorimetry. For a solution with total heat capacity C_soln = 500 J/°C cooling from 26.0°C to 23.0°C, we use q = C × ΔT. With ΔT = 23.0°C - 26.0°C = -3.0°C, we get: q = (500 J/°C)(-3.0°C) = -1500 J = -1.50×10³ J. The negative sign indicates the solution releases heat as it cools. A common error is using the absolute value of ΔT (3.0°C), which would incorrectly give +1.50×10³ J, missing the direction of heat flow. When using total heat capacity, apply q = C × ΔT directly and preserve the sign of ΔT.

Question 3

A 200. g sample of a solution in a coffee-cup calorimeter absorbs +3,600 J+3,600\ \text{J} of heat. The specific heat capacity of the solution is 4.0 J g−1 ∘C−14.0\ \text{J}\,\text{g}^{-1}\,^{\circ}\text{C}^{-1}. What is the temperature change, ΔT\Delta T, of the solution?

  1. +18.0 °C
  2. +4.5 °C (correct answer)
  3. +0.45 °C
  4. +9.0 °C
  5. -4.5 °C
Explanation: This question tests the skill of heat capacity and calorimetry. The temperature change is determined by rearranging q = m c ΔT to ΔT = q / (m c), where q is positive since the solution absorbs heat. The system is the solution, and this equation models the temperature rise due to energy input. With q = +3,600 J, ΔT = 3,600 J / (200 g × 4.0 J g⁻¹ °C⁻¹) = +4.5°C, correctly calculating the change. A tempting distractor is choice A, +18.0°C, which occurs from the misconception of forgetting to include the mass in the denominator. Always rearrange the heat capacity formula carefully and double-check units for consistency.

Question 4

A 40.0 g sample of a solid is heated and its temperature increases by 30.0∘C30.0^{\circ}\text{C}. The specific heat capacity of the solid is 1.2 J g−1 ∘C−11.2\ \text{J}\,\text{g}^{-1}\,^{\circ}\text{C}^{-1}. What is the heat absorbed by the solid, qq?

  1. +1,440 J (correct answer)
  2. +144 J
  3. +3,600 J
  4. +900 J
  5. -1,440 J
Explanation: This question tests the skill of heat capacity and calorimetry. The heat absorbed by the solid is q = m c ΔT, where ΔT is positive for the temperature increase of 30.0°C. The system is the solid, absorbing heat, and the equation models this energy input. Thus, q = 40.0 g × 1.2 J g⁻¹ °C⁻¹ × 30.0°C = +1,440 J, accurately depicting the process. A tempting distractor is choice C, -1,440 J, stemming from the misconception of assigning a negative q to heat absorption. Identify the system and heat direction first, then use q = m c ΔT with appropriate values.

Question 5

A calorimeter has an effective heat capacity of C=80 J/∘CC = 80\ \text{J}/^{\circ}\text{C}. In a trial, the calorimeter releases −1,600 J-1,600\ \text{J} of heat to the surroundings. What is the temperature change, ΔT\Delta T, of the calorimeter?

  1. +20 °C
  2. -20 °C (correct answer)
  3. -12.5 °C
  4. +12.5 °C
  5. -1,280 °C
Explanation: This question tests the skill of heat capacity and calorimetry. The temperature change for the calorimeter is ΔT = q / C, where q is negative because it releases heat. The system is the calorimeter, and this rearranged equation models the cooling due to energy loss. With q = -1,600 J, ΔT = -1,600 J / 80 J °C⁻¹ = -20°C, indicating a temperature decrease. A tempting distractor is choice A, +20°C, which comes from the misconception of ignoring the negative sign of q. Confirm the sign of q based on heat flow direction before solving for ΔT.

Question 6

A student adds 200. g of water to a beaker and supplies heat, causing the water temperature to rise from 15.0°C to 18.0°C. Assume cwater=4.18 J (g⋅∘C)−1c_{\text{water}} = 4.18\ \text{J}\,(\text{g}\cdot{}^\circ\text{C})^{-1}. What is the heat absorbed by the water, qq?

  1. +2.51×10^3 J (correct answer)
  2. +836 J
  3. −2.51×10^3 J
  4. +1.25×10^3 J
  5. +5.02×10^3 J
Explanation: This question tests the skill of heat capacity and calorimetry. The system is the water in the beaker, which absorbs heat, increasing its temperature from 15.0°C to 18.0°C, so ΔT is +3.0°C. The heat absorbed is calculated with q = m c ΔT, using m = 200 g and c = 4.18 J/(g·°C), yielding q = 200 × 4.18 × 3.0 = +2508 J, or +2.51×10^3 J. This equation accurately models the energy transfer as it incorporates the mass-specific response to heat input via specific heat capacity. A tempting distractor is B (+836 J), which results from the misconception of using only one-third of the mass or forgetting to multiply by ΔT fully in q = m c ΔT. Always double-check units and ensure all variables in q = m c ΔT are correctly plugged in for accurate heat calculations.

Question 7

A student places 50.0 g of liquid water in a coffee-cup calorimeter. The temperature of the water increases from 22.0°C to 28.0°C. Assume the water absorbs all the heat released and that the specific heat capacity of water is 4.18 J (g⋅∘C)−14.18\ \text{J}\,(\text{g}\cdot{}^\circ\text{C})^{-1}. What is the heat absorbed by the water, qwaterq_{\text{water}}?

  1. −1.25×10^3 J
  2. +5.02×10^2 J
  3. +2.51×10^3 J
  4. +1.25×10^3 J (correct answer)
  5. −2.51×10^3 J
Explanation: This question tests the skill of heat capacity and calorimetry. The system here is the water in the calorimeter, which absorbs heat, leading to a temperature increase from 22.0°C to 28.0°C, so ΔT is +6.0°C. The heat absorbed by the water is calculated using the equation q = m c ΔT, where m is 50.0 g and c is 4.18 J/(g·°C), resulting in q = 50.0 × 4.18 × 6.0 = +1254 J, or +1.25×10^3 J. This equation models the energy transfer accurately because it accounts for the mass, specific heat, and temperature change, assuming no heat loss to surroundings as stated. A tempting distractor is B (−1.25×103−1.25×10^3 J), which arises from the misconception of assigning a negative sign to heat absorbed, confusing the sign convention where positive q indicates heat gained by the system. To solve similar problems, always calculate ΔT as T_final − T_initial and assign the sign based on whether the system is absorbing or releasing heat.

Question 8

A 100. g100.\ \text{g} sample of liquid ethanol is heated and absorbs 1.00 kJ1.00\ \text{kJ} of heat. The specific heat capacity of ethanol is 2.50 J g−1 °C−12.50\ \text{J}\,\text{g}^{-1}\,\text{°C}^{-1}. What is the temperature change, ΔT\Delta T, of the ethanol?

  1. +0.250\ ^\circ\text{C}
  2. +4.00\ ^\circ\text{C} (correct answer)
  3. +25.0\ ^\circ\text{C}
  4. +2.50\ ^\circ\text{C}
  5. −4.00\ ^\circ\text{C}
Explanation: This problem tests the skill of heat capacity and calorimetry. We need to find ΔT when q, m, and c are known, so we rearrange q = mcΔT to get ΔT = q/(mc). Converting 1.00 kJ to 1000 J, we calculate: ΔT = 1000 J / [(100. g)(2.50 J·g⁻¹·°C⁻¹)] = 1000 / 250 = 4.00°C. Since heat was absorbed (positive q), the temperature increased by 4.00°C. A common mistake would be to select +0.250°C (choice A), which results from forgetting to convert kJ to J before calculating. When solving for ΔT, always ensure your heat units (J or kJ) match the units in the specific heat capacity.

Question 9

A student adds 150. g150.\ \text{g} of water to a calorimeter and heats it from 22.0∘C22.0^\circ\text{C} to 30.0∘C30.0^\circ\text{C}. Assuming the specific heat capacity of water is 4.18 J g−1 °C−14.18\ \text{J}\,\text{g}^{-1}\,\text{°C}^{-1}, what is the heat absorbed by the water, qq, during the heating process?

  1. +5.02\ \text{kJ} (correct answer)
  2. +3.14\ \text{kJ}
  3. −5.02\ \text{kJ}
  4. +1.25\ \text{kJ}
  5. +502\ \text{kJ}
Explanation: This problem tests the skill of heat capacity and calorimetry. The water is being heated from 22.0°C to 30.0°C, so it absorbs heat and q will be positive. Using q = mcΔT, we have q = (150. g)(4.18 J·g⁻¹·°C⁻¹)(30.0°C - 22.0°C) = (150.)(4.18)(8.0) = 5016 J = 5.02 kJ. The positive sign indicates heat absorbed by the water, which makes sense since the temperature increased. A common error would be to use 502 kJ (choice E), which results from forgetting to convert from J to kJ. When calculating heat transfer, always check your units and ensure the sign matches the direction of heat flow (positive for absorption, negative for release).

Question 10

A calorimeter has a heat capacity of C=120 J °C−1C = 120\ \text{J}\,\text{°C}^{-1}. If the calorimeter absorbs 600 J600\ \text{J} of heat, what is the resulting temperature change, ΔT\Delta T, of the calorimeter?

  1. +0.200\ ^\circ\text{C}
  2. +5.00\ ^\circ\text{C} (correct answer)
  3. +720\ ^\circ\text{C}
  4. −5.00\ ^\circ\text{C}
  5. +50.0\ ^\circ\text{C}
Explanation: This problem tests the skill of heat capacity and calorimetry. For a calorimeter with heat capacity C, we use q = CΔT and rearrange to find ΔT = q/C. Given q = 600 J and C = 120 J·°C⁻¹, we calculate: ΔT = 600 J / 120 J·°C⁻¹ = 5.00°C. Since heat was absorbed (positive q), the temperature increased by 5.00°C. A common mistake would be to select +0.200°C (choice A), which results from dividing C by q instead of q by C. When finding temperature change from heat and heat capacity, remember that ΔT = q/C, not C/q.

Question 11

A student cools 200. g200.\ \text{g} of water from 35.0∘C35.0^\circ\text{C} to 25.0∘C25.0^\circ\text{C}. Assume cwater=4.18 J g−1 °C−1c_{\text{water}} = 4.18\ \text{J}\,\text{g}^{-1}\,\text{°C}^{-1}. What is the heat transferred for the water, qwaterq_{\text{water}}?

  1. +8.36\ \text{kJ}
  2. −0.836\ \text{kJ}
  3. −8.36\ \text{kJ} (correct answer)
  4. +0.836\ \text{kJ}
  5. −83.6\ \text{kJ}
Explanation: This problem tests the skill of heat capacity and calorimetry. The water is cooled from 35.0°C to 25.0°C, so it releases heat and q will be negative. Using q = mcΔT: q = (200. g)(4.18 J·g⁻¹·°C⁻¹)(25.0°C - 35.0°C) = (200.)(4.18)(-10.0) = -8360 J = -8.36 kJ. The negative sign correctly indicates heat released by the water as it cooled. A common error would be to choose +8.36 kJ (choice A), forgetting that cooling processes have negative q values. Always determine the sign of q by considering whether the substance gains heat (positive q) or loses heat (negative q) based on the temperature change.

Question 12

A student warms 250. g250.\ \text{g} of an aqueous solution. The solution has a specific heat capacity of 3.80 J g−1 °C−13.80\ \text{J}\,\text{g}^{-1}\,\text{°C}^{-1}, and its temperature increases from 18.0∘C18.0^\circ\text{C} to 24.0∘C24.0^\circ\text{C}. What is the heat absorbed by the solution, qq?

  1. +5.70\ \text{kJ} (correct answer)
  2. +570\ \text{kJ}
  3. −5.70\ \text{kJ}
  4. +3.80\ \text{kJ}
  5. +0.570\ \text{kJ}
Explanation: This problem tests the skill of heat capacity and calorimetry. The solution is warmed from 18.0°C to 24.0°C, so it absorbs heat and q will be positive. Using q = mcΔT: q = (250. g)(3.80 J·g⁻¹·°C⁻¹)(24.0°C - 18.0°C) = (250.)(3.80)(6.0) = 5700 J = 5.70 kJ. The positive value correctly indicates heat absorption during warming. A common error would be to choose +570 kJ (choice B), which results from a decimal error when converting J to kJ. Always double-check unit conversions: 1 kJ = 1000 J, so divide J by 1000 to get kJ.

Question 13

A 75.0 g75.0\ \text{g} sample of aluminum is cooled from 100.0∘C100.0^\circ\text{C} to 40.0∘C40.0^\circ\text{C}. The specific heat capacity of aluminum is 0.900 J g−1 °C−10.900\ \text{J}\,\text{g}^{-1}\,\text{°C}^{-1}. What is the heat transferred for the aluminum sample, qq?

  1. +4.05\ \text{kJ}
  2. −405\ \text{J}
  3. +405\ \text{J}
  4. −4.05\ \text{kJ} (correct answer)
  5. −40.5\ \text{kJ}
Explanation: This problem tests the skill of heat capacity and calorimetry. The aluminum is cooled from 100.0°C to 40.0°C, so it releases heat and q will be negative. Using q = mcΔT: q = (75.0 g)(0.900 J·g⁻¹·°C⁻¹)(40.0°C - 100.0°C) = (75.0)(0.900)(-60.0) = -4050 J = -4.05 kJ. The negative sign correctly indicates heat released during cooling. A common error would be to choose +4.05 kJ (choice A), using the wrong sign by calculating ΔT as (100.0 - 40.0) instead of (final - initial). Always calculate ΔT as (Tfinal - Tinitial) to automatically get the correct sign for q.

Question 14

A 40.0 g40.0\ \text{g} sample of copper absorbs 1.54 kJ1.54\ \text{kJ} of heat. The specific heat capacity of copper is 0.385 J g−1 °C−10.385\ \text{J}\,\text{g}^{-1}\,\text{°C}^{-1}. What is the temperature change, ΔT\Delta T, of the copper?

  1. +10.0\ ^\circ\text{C}
  2. +4.00\ ^\circ\text{C}
  3. +100\ ^\circ\text{C} (correct answer)
  4. −100\ ^\circ\text{C}
  5. +1.00\ ^\circ\text{C}
Explanation: This problem tests the skill of heat capacity and calorimetry. We need to find ΔT using ΔT = q/(mc). First converting 1.54 kJ to 1540 J, we calculate: ΔT = 1540 J / [(40.0 g)(0.385 J·g⁻¹·°C⁻¹)] = 1540 / 15.4 = 100°C. Since heat was absorbed (positive q), the temperature increased by 100°C. A common mistake would be to select +10.0°C (choice A), which results from a calculation error or incorrect unit conversion. When calculating temperature changes from heat absorbed, carefully track your arithmetic and ensure proper unit conversion from kJ to J.

Question 15

A 50.0 g50.0\ \text{g} sample of an unknown metal is warmed from 20.0∘C20.0^\circ\text{C} to 80.0∘C80.0^\circ\text{C}. The metal's specific heat capacity is 0.450 J g−1 °C−10.450\ \text{J}\,\text{g}^{-1}\,\text{°C}^{-1}. What is the heat absorbed by the metal, qq?

  1. +1.35\ \text{kJ} (correct answer)
  2. +13.5\ \text{kJ}
  3. −1.35\ \text{kJ}
  4. +0.675\ \text{kJ}
  5. +2.70\ \text{kJ}
Explanation: This problem tests the skill of heat capacity and calorimetry. The metal is warmed from 20.0°C to 80.0°C, so it absorbs heat and q will be positive. Using q = mcΔT, we calculate q = (50.0 g)(0.450 J·g⁻¹·°C⁻¹)(80.0°C - 20.0°C) = (50.0)(0.450)(60.0) = 1350 J = 1.35 kJ. The positive value correctly indicates heat absorption as the temperature increased. A tempting error would be to select +13.5 kJ (choice B), which results from a decimal place error when converting from J to kJ. To avoid calculation errors, write out all units during the calculation and carefully track decimal places when converting between J and kJ.

Question 16

A coffee-cup calorimeter has a calorimeter constant of C=95 J °C−1C = 95\ \text{J}\,\text{°C}^{-1}. During an experiment, the calorimeter's temperature increases from 24.0∘C24.0^\circ\text{C} to 29.0∘C29.0^\circ\text{C}. What is the heat absorbed by the calorimeter, qcalq_{\text{cal}}?

  1. +475\ \text{J} (correct answer)
  2. −475\ \text{J}
  3. +19\ \text{J}
  4. +95\ \text{J}
  5. +1900\ \text{J}
Explanation: This problem tests the skill of heat capacity and calorimetry. For a calorimeter with a known heat capacity C, we use q = CΔT instead of q = mcΔT. The temperature increases from 24.0°C to 29.0°C, so ΔT = 5.0°C and the calorimeter absorbs heat (positive q). Calculating: q = (95 J·°C⁻¹)(5.0°C) = 475 J, which is positive because heat is absorbed. A common mistake would be to choose -475 J (choice A), incorrectly assigning a negative sign even though the temperature increased. When working with calorimeter constants, remember that q = CΔT directly gives the heat absorbed or released by the calorimeter itself.

Question 17

A coffee-cup calorimeter has a calorimeter constant of Ccal=120 J / ∘CC_{\text{cal}}=120\,\text{J}\,/\,^\circ\text{C}. During a process occurring inside the calorimeter, the temperature of the calorimeter increases from 20.0∘C20.0^\circ\text{C} to 25.0∘C25.0^\circ\text{C}. What is the heat absorbed by the calorimeter, qcalq_{\text{cal}}?

  1. +600 J (correct answer)
  2. −600 J
  3. +24.0 J
  4. +300 J
  5. +720 J
Explanation: This problem tests the skill of heat capacity and calorimetry. For a calorimeter with constant C_cal = 120 J/°C, we use q = C_cal × ΔT instead of q = mcΔT. With ΔT = 25.0°C - 20.0°C = 5.0°C, we calculate: q = (120 J/°C)(5.0°C) = 600 J. Since the calorimeter temperature increases, it absorbs heat, making q positive: +600 J. A common error is confusing the calorimeter constant (J/°C) with specific heat capacity (J/(g·°C)) and trying to use mass in the calculation. Remember that calorimeter constants already incorporate the total heat capacity of the system, so use q = C_cal × ΔT directly.

Question 18

A 200 g200\ \text{g} piece of aluminum is cooled. The specific heat capacity of aluminum is 0.90 J g−1 ∘C−10.90\ \text{J}\,\text{g}^{-1}\,^{\circ}\text{C}^{-1}. If the temperature decreases from 75∘C75^{\circ}\text{C} to 25∘C25^{\circ}\text{C}, what is the heat transferred for the aluminum sample, qq, in joules? (Use the sign convention that heat released by the sample is negative.)

  1. −9.00×10^3 J (correct answer)
  2. +9.00×10^3 J
  3. −1.80×10^2 J
  4. −1.11×10^4 J
  5. −4.50×10^3 J
Explanation: This problem involves heat capacity and calorimetry for a cooling process. When aluminum releases heat during cooling, we apply q = mcΔT with m = 200 g, c = 0.90 J·g⁻¹·°C⁻¹, and ΔT = 25°C - 75°C = -50°C. Calculating: q = (200 g)(0.90 J·g⁻¹·°C⁻¹)(-50°C) = -9,000 J = -9.00×10³ J. The negative sign indicates heat is released by the aluminum as it cools. A common mistake is forgetting the negative sign for cooling processes, which would give +9.00×10³ J (answer B). When solving heat transfer problems, always determine the sign of ΔT first (negative for cooling, positive for heating) to ensure the correct sign for q.

Question 19

A 75.0 g75.0\,\text{g} sample of water cools from 40.0∘C40.0^\circ\text{C} to 30.0∘C30.0^\circ\text{C} in an insulated container. The specific heat capacity of water is 4.18 J (g⋅∘C)−14.18\,\text{J}\,(\text{g}\cdot{}^\circ\text{C})^{-1}. What is the heat transferred for the water, qq?

  1. +314 J
  2. −3.14×10^3 J (correct answer)
  3. −6.27×10^3 J
  4. +3.14×10^3 J
  5. −314 J
Explanation: This problem tests the skill of heat capacity and calorimetry. When water cools from 40.0°C to 30.0°C, we calculate q = mcΔT with m = 75.0 g, c = 4.18 J/(g·°C), and ΔT = 30.0°C - 40.0°C = -10.0°C. Substituting: q = (75.0 g)(4.18 J/(g·°C))(-10.0°C) = -3135 J ≈ -3.14×10³ J. The negative sign indicates heat is released as the water cools. A common mistake is calculating ΔT as +10.0°C (using 40-30 instead of 30-40), which would incorrectly give +3.14×10³ J. For cooling processes, always calculate ΔT = T_final - T_initial to get the correct negative value.

Question 20

A student heats a 100.0 g sample of liquid water in a beaker. The temperature of the water increases from 22.0°C to 28.0°C. Assume the specific heat capacity of water is 4.18 J g−1 ∘C−14.18\ \text{J}\,\text{g}^{-1}\,^{\circ}\text{C}^{-1}. What is the heat transferred to the water, qq?

  1. +2,510 J (correct answer)
  2. +251 J
  3. -2,510 J
  4. +418 J
  5. +10,000 J
Explanation: This question tests the skill of heat capacity and calorimetry. The heat transferred to the water is calculated using the formula q = m c ΔT, where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Here, the system is the water sample, which absorbs heat to increase its temperature from 22.0°C to 28.0°C, resulting in a positive ΔT of 6.0°C. Plugging in the values, q = 100.0 g × 4.18 J g⁻¹ °C⁻¹ × 6.0°C = +2,510 J, correctly modeling the energy absorbed by the water. A tempting distractor is choice C, -2,510 J, which arises from the misconception of assigning a negative sign to q when the system gains heat, confusing the sign convention. Always calculate ΔT as T_final - T_initial and assign the sign of q based on whether the system absorbs (positive) or releases (negative) heat.