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. 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 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.

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.

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.

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.

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.

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.

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.

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.