A 100 g metal block at 80°C is dropped into 100 g of water at 20°C in an insulated cup. A student predicts the final temperature will be 50°C by averaging the two initial temperatures. What is wrong with this prediction?
AThe prediction is correct — equal masses always reach the midpoint temperature
BIt ignores the difference in specific heat capacities; water's specific heat is much higher than most metals, so the final temperature will be closer to 20°C than 50°C
CThe prediction overcounts the heat exchanged because both objects cool simultaneously
DIt is wrong only if the metal undergoes a phase change during cooling
The simple average only works when both objects have identical mass AND specific heat capacity. Water has a specific heat of ~4186 J/kg·K while most metals are much lower (~400 J/kg·K). Because water requires far more energy per degree of temperature change, the equilibrium temperature will be much closer to the water's initial temperature. The correct approach is to set m_metal × c_metal × (T_f − 80) + m_water × c_water × (T_f − 20) = 0 and solve for T_f.
Question 2 Multiple Choice
A student dissolves a salt in water inside a coffee-cup calorimeter and observes that the temperature of the solution drops by 4°C. Which conclusion is correct?
AThe dissolution is exothermic; heat flowed from the solution into the salt
BThe dissolution is endothermic; the salt absorbed heat from the water, cooling the solution
CThe temperature drop indicates a violation of conservation of energy
DThe result is inconclusive because coffee-cup calorimeters only work for temperature increases
A temperature drop in the solution means the solution lost thermal energy — it was the 'hot' object that gave up heat. The salt absorbed that heat during dissolution, making the process endothermic (Q_solution < 0, so Q_dissolution > 0). Conservation of energy still holds: heat lost by the solution equals heat gained by the dissolution process. This is exactly how instant cold packs work — ammonium nitrate dissolving in water is endothermic.
Question 3 True / False
In the standard calorimetry equation Q_lost + Q_gained = 0, the hot object's Q term is negative because its temperature decreases, giving a negative ΔT in Q = mcΔT.
TTrue
FFalse
Answer: True
This is the key to correct sign conventions. Q = mcΔT where ΔT = T_final − T_initial. For the hot object, T_final < T_initial, so ΔT < 0 and Q < 0 — it lost heat. For the cool object, T_final > T_initial, so ΔT > 0 and Q > 0 — it gained heat. Setting up the equation this way (rather than writing |Q_lost| = |Q_gained|) keeps the signs consistent and automatically satisfies conservation of energy.
Question 4 True / False
A bomb calorimeter is used to measure the enthalpy change ΔH of a combustion reaction.
TTrue
FFalse
Answer: False
A bomb calorimeter is a sealed, rigid vessel — it operates at constant volume, so it measures the change in internal energy ΔU, not enthalpy ΔH. It is the coffee-cup calorimeter (open to the atmosphere, constant pressure) that approximates ΔH. For reactions involving only liquids and solids the difference is small, but for reactions producing gases, ΔH = ΔU + ΔnRT, where Δn is the change in moles of gas. Confusing the two calorimeter types leads to reporting the wrong thermodynamic quantity.
Question 5 Short Answer
Why is the final equilibrium temperature in a calorimetry experiment not simply the arithmetic average of the two initial temperatures, even when the masses of the two objects are equal?
Think about your answer, then reveal below.
Model answer: Because equilibrium temperature depends on both mass and specific heat capacity. The equation Q_lost + Q_gained = 0 expands to m₁c₁(T_f − T₁) + m₂c₂(T_f − T₂) = 0. If the specific heat capacities c₁ and c₂ differ, T_f is a weighted average biased toward the substance with higher heat capacity, not a simple midpoint. For example, water has a much higher specific heat than iron, so mixing equal masses of hot iron and cool water gives a final temperature much closer to the water's initial temperature.
The arithmetic average assumes each degree of temperature change transfers the same amount of heat for both objects, which is only true when mc is identical for both. Specific heat capacity quantifies how much energy is stored per gram per degree — water stores roughly ten times as much as iron. This is why oceans moderate coastal climates: water's high specific heat means it absorbs a huge amount of energy for a small temperature change.