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.

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.

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.

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.

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.

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.

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.

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.

This question tests the skill of heat capacity and calorimetry. The heat absorbed by the aluminum is q = m c ΔT, with positive ΔT since temperature increases from 20.0°C to 45.0°C. The system is the aluminum sample, gaining heat, and the equation models this endothermic process. Thus, q = 120 g × 0.90 J g⁻¹ °C⁻¹ × 25°C = +2,700 J, correctly quantifying the energy transfer. A tempting distractor is choice C, -2,700 J, arising from the misconception of using a negative sign for q when the system absorbs heat. Start by noting if the temperature increases (positive q) or decreases (negative q), then apply q = m c ΔT accordingly.

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.