Heat Capacity and Calorimetry - AP Chemistry

1. A metal block with mass 200. g has a specific heat capacity of $0.500\ \text{J}\,(\text{g}\cdot{}^\circ\text{C})^{-1}$. If the block absorbs $+1.50\times 10^3\ \text{J}$ of heat, what is the temperature change, $\Delta T$, of the block?
   Answer: +15.0 °C This question tests the skill of heat capacity and calorimetry. The system is the metal block, which absorbs +1.50×$10^3$ J of heat, leading to a temperature increase. The temperature change is found by rearranging q = m c ΔT to ΔT = q / (m c), with m = 200 g and c = 0.500 J/(g·°C), giving ΔT = 1500 / (200 × 0.500) = +15.0°C. This rearranged equation models the energy transfer by solving for the effect of known heat input on temperature. A tempting distractor is A (+30.0 °C), resulting from the misconception of forgetting to multiply by c in the denominator. When solving for ΔT, always include all components of q = m c ΔT in the rearrangement.
2. A $40.0\,\text{g}$ sample of aluminum is heated, and its temperature increases from $25.0^\circ\text{C}$ to $50.0^\circ\text{C}$. The specific heat capacity of aluminum is $0.900\,\text{J}\,(\text{g}\cdot{}^\circ\text{C})^{-1}$. What is the heat absorbed by the aluminum sample, $q$?
   Answer: +900 J 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.
3. 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 $4.18\ \text{J}\,(\text{g}\cdot{}^\circ\text{C})^{-1}$. What is the heat absorbed by the water, $q_{\text{water}}$?
   Answer: +1.25×$10^3$ J 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 ($−1.25×10^3$ 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.
4. A 200. g sample of a solution in a coffee-cup calorimeter absorbs $+3,600\ \text{J}$ of heat. The specific heat capacity of the solution is $4.0\ \text{J}\,\text{g}^{-1}\,^{\circ}\text{C}^{-1}$. What is the temperature change, $\Delta T$, of the solution?
   Answer: +4.5 °C 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.
5. A coffee-cup calorimeter has a calorimeter constant of $C_{\text{cal}}=120\ \text{J}\,^{\circ}\text{C}^{-1}$. During an experiment, the calorimeter temperature increases from $20.0^{\circ}\text{C}$ to $24.0^{\circ}\text{C}$. What heat is absorbed by the calorimeter, $q_{\text{cal}}$, in joules?
   Answer: +480 J This question tests heat capacity and calorimetry using a calorimeter constant. For a calorimeter with constant C_cal = 120 J·°C⁻¹, the heat absorbed is q_cal = C_cal × ΔT, where ΔT = 24.0°C - 20.0°C = 4.0°C. Therefore, q_cal = (120 J·°C⁻¹)(4.0°C) = 480 J. Since the temperature increases, the calorimeter absorbs heat, making q positive: +480 J. A common error is confusing the sign convention and reporting -480 J (answer B), thinking that the calorimeter must release heat. Remember that when using calorimeter constants, positive ΔT means heat absorbed by the calorimeter, resulting in positive q_cal.
6. A 40.0 g sample of a solid is heated and its temperature increases by $30.0^{\circ}\text{C}$. The specific heat capacity of the solid is $1.2\ \text{J}\,\text{g}^{-1}\,^{\circ}\text{C}^{-1}$. What is the heat absorbed by the solid, $q$?
   Answer: +1,440 J 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.
7. 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 $c_{\text{water}} = 4.18\ \text{J}\,(\text{g}\cdot{}^\circ\text{C})^{-1}$. What is the heat absorbed by the water, $q$?
   Answer: +2.51×$10^3$ J 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.
8. A student cools $200.\ \text{g}$ of water from $35.0^\circ\text{C}$ to $25.0^\circ\text{C}$. Assume $c_{\text{water}} = 4.18\ \text{J}\,\text{g}^{-1}\,\text{°C}^{-1}$. What is the heat transferred for the water, $q_{\text{water}}$?
   Answer: −8.36\ \text{kJ} 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.
9. A student heats a $150\ \text{g}$ sample of liquid water in a cup. The specific heat capacity of water is $4.18\ \text{J}\,\text{g}^{-1}\,^{\circ}\text{C}^{-1}$. If the temperature of the water increases from $22.0^{\circ}\text{C}$ to $30.0^{\circ}\text{C}$, what is the heat absorbed by the water, $q$, in joules?
   Answer: +5.02×$10^3$ J This problem tests your understanding of heat capacity and calorimetry. When water absorbs heat, we use the equation q = mcΔT, where q is heat absorbed, m is mass (150 g), c is specific heat capacity (4.18 J·g⁻¹·°C⁻¹), and ΔT is the temperature change (30.0°C - 22.0°C = 8.0°C). Substituting these values: q = (150 g)(4.18 J·g⁻¹·°C⁻¹)(8.0°C) = 5,016 J ≈ 5.02×10³ J. Since the water absorbs heat (temperature increases), q is positive. A common error would be using the wrong temperature change, such as calculating 30.0°C + 22.0°C = 52.0°C instead of the difference, which would give answer E. To solve calorimetry problems correctly, always calculate ΔT as (final temperature - initial temperature) and assign the correct sign based on whether heat is absorbed (+) or released (-).
10. A $300\ \text{g}$ sample of ethanol is warmed. The specific heat capacity of ethanol is $2.4\ \text{J}\,\text{g}^{-1}\,^{\circ}\text{C}^{-1}$. If the sample absorbs $3.60\times10^3\ \text{J}$ of heat, what is the final temperature of the ethanol if it starts at $15.0^{\circ}\text{C}$? (Assume no heat loss.)
    Answer: 20.0 °C This problem tests heat capacity and calorimetry to find final temperature. First, calculate the temperature change using q = mcΔT: 3.60×10³ J = (300 g)(2.4 J·g⁻¹·°C⁻¹)(ΔT). Solving: ΔT = 3,600 J/(720 J·°C⁻¹) = 5.0°C. Since the ethanol absorbs heat, this is a temperature increase: $T_f$ = $T_i$ + ΔT = 15.0°C + 5.0°C = 20.0°C. A common mistake is adding the heat directly to the initial temperature without converting to temperature change, or using the wrong mass, which might give 25.0°C (answer C). To find final temperature after heating, always calculate ΔT = q/(mc) first, then add to the initial temperature.
11. A coffee-cup calorimeter has a calorimeter constant of $C_{\text{cal}}=120\,\text{J}\,/\,^\circ\text{C}$. During a process occurring inside the calorimeter, the temperature of the calorimeter increases from $20.0^\circ\text{C}$ to $25.0^\circ\text{C}$. What is the heat absorbed by the calorimeter, $q_{\text{cal}}$?
    Answer: +600 J 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.
12. A $200\ \text{g}$ piece of aluminum is cooled. The specific heat capacity of aluminum is $0.90\ \text{J}\,\text{g}^{-1}\,^{\circ}\text{C}^{-1}$. If the temperature decreases from $75^{\circ}\text{C}$ to $25^{\circ}\text{C}$, what is the heat transferred for the aluminum sample, $q$, in joules? (Use the sign convention that heat released by the sample is negative.)
    Answer: −9.00×$10^3$ J 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.
13. In an insulated calorimeter, the heat gained by the calorimeter is equal in magnitude to the heat lost by a hot object placed inside it. The calorimeter constant is $C_{\text{cal}} = 200\ \text{J}/^{\circ}\text{C}$, and the calorimeter temperature increases from 19.0°C to 25.0°C. How much heat is absorbed by the calorimeter, $q_{\text{cal}}$?
    Answer: +1,200 J This question tests the skill of heat capacity and calorimetry. The heat absorbed by the calorimeter is q_cal = C_cal × ΔT, with ΔT positive as temperature rises from 19.0°C to 25.0°C. The system is the calorimeter, gaining heat from the hot object, and the equation models this transfer. With ΔT = 6.0°C, q_cal = 200 J °C⁻¹ × 6.0°C = +1,200 J, correctly calculating the absorption. A tempting distractor is choice A, -1,200 J, which arises from the misconception of using a negative sign for heat gained by the calorimeter. In calorimetry problems, equate heat lost by one part to heat gained by another, ensuring consistent signs.
14. A 60.0 g sample of a liquid cools from 40.0°C to 20.0°C. The specific heat capacity of the liquid is $3.00\ \text{J}\,(\text{g}\cdot{}^\circ\text{C})^{-1}$. What is the heat transferred for the liquid, $q$ (sign included)?
    Answer: −3.60×$10^3$ J This question tests the skill of heat capacity and calorimetry. The system is the liquid sample, which releases heat as it cools from 40.0°C to 20.0°C, so ΔT is −20.0°C. The heat transferred is q = m c ΔT, with m = 60.0 g and c = 3.00 J/(g·°C), yielding q = 60.0 × 3.00 × (−20.0) = −3600 J, or −3.60×$10^3$ J. This equation accurately models the energy transfer with the negative sign indicating heat loss from the system. A tempting distractor is A ($+3.60×10^3$ J), which arises from the misconception of ignoring the sign of ΔT and treating cooling as positive. Remember to use the sign of ΔT to determine if q is positive (absorption) or negative (release).
15. A student heats a 50.0 g sample of liquid water in a beaker. The specific heat capacity of water is $4.18\ \text{J}\,\text{g}^{-1}\,^{\circ}\text{C}^{-1}$. The temperature of the water increases from $22.0^{\circ}\text{C}$ to $30.0^{\circ}\text{C}$. What is the heat absorbed by the water, $q$, in joules?
    Answer: +1.67×$10^3$ J This problem tests your understanding of heat capacity and calorimetry. When water is heated, it absorbs energy according to q = mcΔT, where m is mass (50.0 g), c is specific heat capacity (4.18 J/g°C), and ΔT is the temperature change (30.0°C - 22.0°C = 8.0°C). Substituting these values: q = (50.0 g)(4.18 J/g°C)(8.0°C) = 1,672 J = 1.67×10³ J. Since the water temperature increases, heat is absorbed, making q positive. A common error would be using the final temperature (30.0°C) instead of the temperature change (8.0°C), which would give 6,270 J. To solve calorimetry problems correctly, always calculate ΔT = Tfinal - Tinitial before applying q = mcΔT.
16. A calorimeter has a total heat capacity of $C = 250\ \text{J}\,^{\circ}\text{C}^{-1}$. If the calorimeter releases $q = -1.00\times10^3\ \text{J}$ of heat, what is the temperature change, $\Delta T$, in $^{\circ}\text{C}$?
    Answer: −4.00 °C This problem involves heat capacity and calorimetry using a calorimeter's total heat capacity. Given that the calorimeter releases q = -1.00×10³ J, we use ΔT = q/C where C = 250 J/°C. Calculating: ΔT = (-1.00×10³ J)/(250 J/°C) = -1,000 J / 250 J/°C = -4.00°C. The negative temperature change confirms the calorimeter cooled as it released heat. A common mistake would be getting -2.50°C by incorrectly dividing 1,000 by 400 instead of 250. When using calorimeter heat capacity C, remember it already combines mass and specific heat, so divide q directly by C.
17. A student adds $q = +900\ \text{J}$ of heat to a metal sample. The sample has a mass of 100 g and a specific heat capacity of $0.300\ \text{J}\,\text{g}^{-1}\,^{\circ}\text{C}^{-1}$. What is the temperature change, $\Delta T$, in $^{\circ}\text{C}$?
    Answer: +30 °C This problem involves heat capacity and calorimetry to find temperature change. Given q = +900 J added to the metal, we use ΔT = q/(mc) where m = 100 g and c = 0.300 J/g°C. Calculating: ΔT = (900 J)/[(100 g)(0.300 J/g°C)] = 900 J / 30 J/°C = 30°C. Since heat is added (positive q), the temperature increases by 30°C. A common mistake is dividing 900 by 300 to get 3.0°C, forgetting to include the mass in the calculation. When finding temperature change, always divide q by the product of mass and specific heat capacity.
18. A calorimeter has a calorimeter constant of $C_{\text{cal}} = 250\ \text{J}/^\circ\text{C}$. If the calorimeter releases $-1.00\times 10^3\ \text{J}$ of heat, what is the temperature change of the calorimeter, $\Delta T$?
    Answer: −4.0 °C This question tests the skill of heat capacity and calorimetry. The system is the calorimeter, which releases −1.00×$10^3$ J of heat, causing a temperature decrease. The temperature change is calculated by ΔT = q / C_cal, using C_cal = 250 J/°C, resulting in ΔT = −1000 / 250 = −4.0°C. This equation models the energy transfer by relating the heat released directly to the calorimeter's capacity constant. A tempting distractor is B (−2.5 °C), stemming from the misconception of using an incorrect divisor or miscalculating the division. For calorimeter constants, treat them like heat capacities and solve for ΔT with proper sign consideration.
19. A $100\ \text{g}$ sample of water releases $4.18\times10^3\ \text{J}$ of heat as it cools. The specific heat capacity of water is $4.18\ \text{J}\,\text{g}^{-1}\,^{\circ}\text{C}^{-1}$. If the initial temperature is $35.0^{\circ}\text{C}$, what is the final temperature of the water? (Assume no heat loss to the surroundings.)
    Answer: 25.0 °C This problem involves heat capacity and calorimetry to find final temperature. When water releases 4.18×10³ J of heat, q = -4.18×10³ J (negative for heat released). Using q = mcΔT with m = 100 g and c = 4.18 J·g⁻¹·°C⁻¹: -4,180 J = (100 g)(4.18 J·g⁻¹·°C⁻¹)($T_f$ - 35.0°C). Solving: -4,180 = 418($T_f$ - 35.0), so $T_f$ - 35.0 = -10.0°C, giving $T_f$ = 25.0°C. A common error is forgetting the negative sign for released heat, which would give $T_f$ = 45.0°C (answer B). When heat is released, always use negative q and solve for the final temperature algebraically.
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 $4.18\ \text{J}\,\text{g}^{-1}\,^{\circ}\text{C}^{-1}$. What is the heat transferred to the water, $q$?
    Answer: +2,510 J 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.

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