7 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 7
A 50.0 g block of copper at 80.0°C is placed in contact with a 200.0 g block of aluminum at 30.0°C inside an insulated box. Which statement describes the energy transfer that occurs?
Answer and reasoning
AEnergy is transferred from the aluminum to the copper until both contain equal thermal energy A student who thinks heat flows from the object containing more thermal energy, here the larger aluminum block, to the one containing less picks this. The direction is set by temperature: the copper is hotter, so energy flows from the copper to the aluminum, and it stops when the temperatures are equal.
BCold is transferred from the aluminum to the copper until their temperatures are equal A student who thinks cold flows from a cold object into a warm one picks this. Cold is not transferred; the copper cools because it loses energy to the aluminum.
CEnergy is transferred from the copper to the aluminum until their temperatures are equalCorrect Heat flows from the body at the higher temperature to the body at the lower temperature. Collisions at the contact surface transfer energy from the copper to the aluminum until the particles in both blocks have the same average kinetic energy, that is, until their temperatures are equal.
DEnergy is transferred from the copper to the aluminum until both hold equal thermal energy A student who thinks thermal equilibrium means equal amounts of thermal energy picks this. Transfer stops when the temperatures are equal; at that point the larger aluminum block holds more total thermal energy than the copper.
A 250.0 g sample of an unknown metal at 100.0°C is placed in 50.0 g of water at 20.0°C in an insulated calorimeter of negligible heat capacity. The metal and the water reach thermal equilibrium at 30.0°C. The specific heat capacity of water is 4.18 J/(g·°C). What is the specific heat capacity of the metal?
Answer and reasoning
A0.597 J/(g·°C) A student who uses the mass of the metal, the substance being studied, with the water's specific heat capacity and temperature change picks this: (250.0)(4.18)(10.0) J divided by (250.0 g)(70.0°C). The heat absorbed by the water must be calculated with the water's mass, 50.0 g.
B0.836 J/(g·°C) A student who uses the temperature reached, 30.0°C, as the temperature change of both the water and the metal picks this: (50.0)(4.18)(30.0)/((250.0)(30.0)). ΔT is the difference between the final and initial temperatures: 10.0°C for the water and 70.0°C for the metal.
C0.119 J/(g·°C)Correct The water absorbed (50.0 g)(4.18 J/(g·°C))(10.0°C) = 2.09 × 10³ J. Because energy is conserved, the metal released the same amount while cooling by 70.0°C, so c = (2.09 × 10³ J)/((250.0 g)(70.0°C)) = 0.119 J/(g·°C).
D0.690 J/(g·°C) A student who adds 273 to each temperature change to convert it to kelvins picks this: (50.0)(4.18)(283.0)/((250.0)(343.0)). A temperature change has the same size in kelvins as in degrees Celsius, so nothing is added.
Working Heat absorbed by water: q = (50.0 g)(4.18 J/(g·°C))(30.0°C − 20.0°C) = 2.09 × 10³ J. Energy is conserved, so the metal released 2.09 × 10³ J while cooling by 100.0°C − 30.0°C = 70.0°C. c = q/(mΔT) = (2.09 × 10³ J)/((250.0 g)(70.0°C)) = 0.119 J/(g·°C).
Equal masses of liquid water and liquid ethanol absorb equal amounts of energy, and neither liquid boils. The temperature of the ethanol rises by 12.0°C. Based on the data in the table, by how much does the temperature of the water rise?
Answer and reasoning
A20.6°C A student who thinks a larger specific heat capacity gives a larger temperature change picks this, multiplying 12.0°C by 4.18/2.44. Because ΔT = q/(mc), the liquid with the larger c, water, has the smaller rise.
B12.0°C A student who thinks equal heats give equal temperature changes in equal masses of any substance picks this. The two liquids have different specific heat capacities, so their temperature changes differ.
C7.00°CCorrect With q and m the same for both liquids, ΔT is inversely proportional to the specific heat capacity: ΔTwater = 12.0°C × (2.44/4.18) = 7.00°C. Water's larger specific heat capacity gives it the smaller temperature rise.
D17.9°C A student who thinks equal masses contain equal amounts in moles picks this, comparing molar heat capacities: 12.0°C × (112.4/75.3). Equal masses of water and ethanol contain different numbers of moles, so the specific heat capacities are the ones to compare.
Working q = mcΔT with q and m equal for both, so ΔT ∝ 1/c: ΔTwater = 12.0°C × (2.44/4.18) = 7.00°C.
The box labeled Before represents a sample of N₂(g) in a sealed, rigid container at 25°C. The line behind each molecule indicates its speed; a longer line means a faster molecule. The sample is heated to 75°C. Which of the numbered diagrams best represents the sample at 75°C?
Answer and reasoning
ADiagram 1 A student who thinks particles expand when heated picks this. The molecules themselves do not change size; heating makes them move faster, which this diagram does not show.
BDiagram 2 A student who thinks heated gas particles rise to the top of a container picks this. In a sealed container heated evenly, the molecules stay spread through the whole volume and move faster.
CDiagram 3 A student who thinks heating breaks molecules into atoms picks this. Heating N₂ from 25°C to 75°C makes the molecules move faster; it does not break the bond between the two N atoms.
DDiagram 4Correct Heating the gas increases its energy: the molecules move faster on average, shown by the longer lines. The molecules keep their size and identity and stay spread through the whole container.
The molar heat capacity of aluminum is 24.2 J/(mol·°C). How much heat must be absorbed to raise the temperature of 40.0 g of aluminum from 20.0°C to 55.0°C?
Answer and reasoning
A3.39 × 10⁴ J A student who treats a molar heat capacity as if it were a specific heat capacity picks this, multiplying 24.2 J/(mol·°C) by the mass in grams. The value is per mole, so the mass must first be converted to moles.
B9.14 × 10⁵ J A student who multiplies the mass by the molar mass to find the moles picks this, using 40.0 × 26.98 = 1079 'mol'. The amount is the mass divided by the molar mass, 1.48 mol.
C1.11 × 10⁴ J A student who adds 273 to the temperature change picks this, using ΔT = 308. A rise of 35.0°C is a rise of 35.0 K, so ΔT = 35.0.
D1.26 × 10³ JCorrect n = 40.0 g ÷ 26.98 g/mol = 1.48 mol, and ΔT = 35.0°C. q = n × Cmolar × ΔT = (1.48 mol)(24.2 J/(mol·°C))(35.0°C) = 1.26 × 10³ J.
Working n = 40.0 g ÷ 26.98 g/mol = 1.48 mol. ΔT = 55.0°C − 20.0°C = 35.0°C. q = n × Cmolar × ΔT = (1.48 mol)(24.2 J/(mol·°C))(35.0°C) = 1.26 × 10³ J.
In which of the following does the energy of the system change because of a chemical reaction, rather than because of heating or cooling or a phase transition?
Answer and reasoning
ALiquid wax solidifying as it drips down a candle A student who thinks a change of state is a chemical reaction picks this. Solidified wax is still wax; the energy released comes from a phase transition.
BA tungsten filament glowing inside a light bulb A student who thinks glowing always shows a chemical reaction picks this. The filament is heated by the electric current and glows because it is very hot; it is still tungsten, so its energy changes by heating.
CParaffin wax vapor burning in a candle flameCorrect In the flame, wax vapor reacts with oxygen to form new substances, carbon dioxide and water, and releases energy. The energy change comes from a chemical reaction.
DWater forming on the outside of a glass of ice water A student who thinks the cold glass makes water out of air picks this. Water vapor already in the air condenses on the cold surface, a phase transition.
When 10.0 g of solid X dissolves in 40.0 g of water at 21.0°C in an insulated calorimeter, the dissolution releases 1.67 kJ of energy. Assume that the specific heat capacity of the solution is 4.18 J/(g·°C) and that no energy is lost from the calorimeter. What is the final temperature of the solution?
Answer and reasoning
A31.0°C A student who thinks a dissolved solid no longer counts toward the solution's mass picks this, using 40.0 g. The solution's mass is 50.0 g, so the rise is 7.99°C, not 9.99°C.
B29.0°CCorrect The solution's mass is 10.0 g + 40.0 g = 50.0 g. It absorbs the 1.67 kJ released, so ΔT = 1670 J ÷ ((50.0 g)(4.18 J/(g·°C))) = 7.99°C, and the final temperature is 21.0°C + 7.99°C = 29.0°C.
C61.0°C A student who uses the mass of the dissolving solid in q = mcΔT picks this, giving a rise of 40.0°C. The energy is absorbed by all 50.0 g of solution, whose temperature is measured.
D13.0°C A student who thinks a process that releases energy makes the thermometer reading fall picks this. The energy released by the dissolution is absorbed by the solution, so the solution's temperature rises.
Working Mass of solution = 10.0 g + 40.0 g = 50.0 g. The solution absorbs the 1.67 kJ released: ΔT = q/(mc) = 1670 J ÷ ((50.0 g)(4.18 J/(g·°C))) = 7.99°C. Tfinal = 21.0°C + 7.99°C = 29.0°C.
In preparation: 0 of 7 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
6.4.A.1 Heat (q) Fix
Heat (q)
Energy transferred between two bodies because of a difference in temperature. Heat flows from the warmer body to the cooler body until the two reach the same temperature.
Heat transfer equation, q = mcΔT
The heat absorbed or released by a sample that is heated or cooled without a phase change or reaction: q is the heat, m the mass, c the specific heat capacity and ΔT = Tfinal − Tinitial. q is positive when the sample absorbs heat and negative when it releases heat.
Calorimetry
Measuring heat transfer from the temperature change of a known mass of a substance, usually water or an aqueous solution, held in an insulated container (a calorimeter). The heat gained or lost by the calorimeter contents is found with q = mcΔT.
Temperature change, ΔT
Tfinal − Tinitial: positive when a sample warms and negative when it cools. A kelvin and a degree Celsius are the same size, so a temperature change has the same value in K and in °C.
Students often think A temperature change must be converted to kelvins by adding 273 to it, so ΔT in kelvins is 273 more than ΔT in degrees Celsius. In fact No. A kelvin and a degree Celsius are the same size, so a temperature change has the same value in K and in °C: a rise from 18.0°C to 42.0°C is a rise of 24.0 K. Adding 273 converts a temperature, not a difference between two temperatures.
Students often think The heat needed for a given temperature change depends only on the substance and the temperature change, not on how much of the substance is present. In fact Yes. q = mcΔT, so the heat needed is proportional to the mass: twice the mass of the same substance needs twice the heat for the same temperature change, and the same heat gives it half the temperature change.
6.4.A.2 First law of thermodynamics Fix
First law of thermodynamics
Energy is conserved in chemical and physical processes: energy lost by one part of a system is gained by another part or by the surroundings, and none is created or destroyed.
Heat exchange in an insulated calorimeter
When no energy enters or leaves the calorimeter, the heat released by the warmer body equals the heat absorbed by the cooler body: qwarm + qcool = 0. The two temperature changes differ unless the bodies have equal values of m × c.
Students often think The final temperature of two objects brought into thermal contact is the average of their starting temperatures, whatever their masses and specific heat capacities. In fact Only when the two objects have equal values of m × c, such as equal masses of water. Otherwise the final temperature lies closer to the starting temperature of the object with the larger m × c, and it is found from heat lost = heat gained.
Students often think Heat and temperature are the same thing, so the object whose temperature changes more must have gained or lost more heat. In fact No. Temperature describes how hot an object is; heat is energy transferred. The heat an object gains or loses also depends on its mass and specific heat capacity (q = mcΔT), so in an insulated calorimeter a hot metal can cool by 65.0°C while water warms by 5.0°C with equal amounts of heat transferred.
6.4.A.3 Specific heat capacity (c) Fix
Specific heat capacity (c)
The energy needed to raise the temperature of 1 g of a substance by 1°C, in J/(g·°C); for water, 4.18 J/(g·°C). For the same heat and the same mass, a substance with a smaller c undergoes a larger temperature change.
Students often think A substance with a larger specific heat capacity undergoes a larger temperature change for the same amount of heat, as if ΔT were proportional to c. In fact No. A larger specific heat capacity means more energy is needed per gram for each degree, so for the same heat and mass a substance with a larger c undergoes a smaller temperature change.
Students often think The same amount of heat produces the same temperature change in equal masses of any substance; only the heat and the mass matter, not which substance it is. In fact No. The temperature change also depends on the specific heat capacity, ΔT = q/(mc), so for equal heats and equal masses the substance with the smaller specific heat capacity undergoes the larger temperature change.
6.4.A.4 Energy change on heating and cooling Fix
Energy change on heating and cooling
Heating a system increases its energy (q > 0) and cooling it decreases its energy (q < 0). For a sample that is not changing phase or reacting, heating raises the average kinetic energy of its particles, so its temperature rises.
Students often think When a substance is heated its particles expand, which is why the substance itself expands. In fact No. Heating increases the average kinetic energy of the molecules, so they move faster on average; the size of each molecule does not change.
Students often think When a gas is heated its particles rise to the top of the container, because hot gas rises. In fact No. The molecules move faster on average and remain spread through the whole container; a sealed, rigid container holds the same number of molecules in the same volume after heating.
6.4.A.5 Molar heat capacity Fix
Molar heat capacity
The energy needed to raise the temperature of 1 mol of a substance by 1°C, in J/(mol·°C). It is used with an amount in moles (q = n × Cmolar × ΔT) and equals the specific heat capacity multiplied by the molar mass.
Students often think Equal masses of two substances contain equal amounts in moles, so their molar heat capacities can be compared directly when the masses are equal. In fact No. The amount in moles is the mass divided by the molar mass, so equal masses of water (18.02 g/mol) and ethanol (46.07 g/mol) contain different amounts: 100. g of water is 5.55 mol, and 100. g of ethanol is 2.17 mol.
Students often think A specific heat capacity and a molar heat capacity are the same quantity, so either value can be used with a mass in grams and no conversion between them is needed. In fact No. A specific heat capacity, in J/(g·°C), is multiplied by a mass in grams; a molar heat capacity, in J/(mol·°C), is multiplied by an amount in moles. For a given substance the two values differ by a factor equal to its molar mass.
6.4.A.6 Three ways a chemical system changes its energy Fix
Three ways a chemical system changes its energy
Heating or cooling, phase transitions (such as melting, freezing, boiling or condensing) and chemical reactions. Ice melting at 0°C absorbs energy even though its temperature does not change.
Students often think A change of state, such as wax solidifying or ice melting, is a chemical reaction because the substance looks different afterward. In fact No. In a phase change the substance stays the same; solid wax and liquid wax are the same compounds, and only the arrangement and motion of the particles change. A chemical reaction forms new substances.
Students often think Glowing or a change in color always shows that a chemical reaction is taking place. In fact No. A very hot solid gives off light because of its high temperature; a tungsten filament glowing in a light bulb is still tungsten. A color change or light is evidence of a reaction only when new substances form.
6.4.A.7 Exothermic and endothermic dissolution in a calorimeter Fix
Exothermic and endothermic dissolution in a calorimeter
If the temperature of the mixture rises as a solute dissolves, the dissolution process released thermal energy (exothermic); if the temperature falls, the dissolution process absorbed thermal energy (endothermic).
System and surroundings in solution calorimetry
The dissolving solute (the process) is the system; the water or solution whose temperature is measured is the surroundings. The heat for the process has the same size as, and the opposite sign to, the heat for the solution.
Students often think A solid that dissolves no longer adds to the mass of the solution, because its particles disappear into the water, so the solution's mass is just the mass of the water. In fact No. Mass is conserved when a solid dissolves: the mass of the solution equals the mass of the water plus the mass of the dissolved solid, and all of that mass changes temperature in the calorimeter.
Students often think The thermometer reading describes the dissolving substance itself: a rise in temperature means the dissolving process absorbed energy, and a fall means it released energy. In fact No. The thermometer measures the water or solution, which is the surroundings of the dissolving process. If the process releases energy, the solution gains it and its temperature rises (exothermic); if the process absorbs energy, it takes it from the solution and the temperature falls (endothermic).
11 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 11
A 250.0 g sample of water in an insulated container is warmed by an electric heater from 18.0°C to 42.0°C. The specific heat capacity of water is 4.18 J/(g·°C). How much heat did the water absorb?
Answer and reasoning
A2.51 × 10⁴ JCorrect ΔT = 42.0°C − 18.0°C = 24.0°C, so q = mcΔT = (250.0 g)(4.18 J/(g·°C))(24.0°C) = 2.51 × 10⁴ J. The water absorbs heat, so q is positive.
B3.10 × 10⁵ J A student who thinks a temperature change must be converted to kelvins by adding 273 picks this, using ΔT = 297. A rise of 24.0°C is a rise of 24.0 K, so ΔT = 24.0.
C1.00 × 10² J A student who thinks the heat needed for a temperature change does not depend on the amount of water picks this: cΔT = 4.18 × 24.0 = 1.00 × 10² J is the heat for only 1 g. The 250.0 g sample needs 250.0 times as much.
D4.39 × 10⁴ J A student who uses the temperature reached instead of the temperature change picks this, calculating (250.0)(4.18)(42.0). q = mcΔT needs ΔT = 42.0°C − 18.0°C = 24.0°C.
Working ΔT = 42.0°C − 18.0°C = 24.0°C. q = mcΔT = (250.0 g)(4.18 J/(g·°C))(24.0°C) = 2.51 × 10⁴ J (positive: the water absorbs heat).
A student dissolves a solid in water in a foam-cup calorimeter and records the data shown in the table. The student will calculate the heat gained or lost by the solution with q = mcΔT. Based on the data, what mass should the student use for m?
Answer and reasoning
A50.08 g A student who thinks a dissolved solid no longer adds to the mass of the solution picks this, using the water alone (53.20 g − 3.12 g). Mass is conserved when the solid dissolves, so the solution also contains the 10.50 g of solid.
B10.50 g A student who thinks m is the mass of the substance being studied picks the mass of the solid (63.70 g − 53.20 g). The temperature change measured is that of the whole solution, so m must be the solution's mass.
C63.70 g A student who counts the cup in the mass used with the solution's specific heat capacity picks the total balance reading. The cup is not part of the solution, so its 3.12 g must be subtracted.
D60.58 gCorrect The solution is the water plus the dissolved solid, and its mass is found by difference: 63.70 g − 3.12 g = 60.58 g. This is the mass whose temperature changed, so it is used with the solution's specific heat capacity.
Working Mass of solution = (mass of cup + water + dissolved solid) − (mass of empty cup) = 63.70 g − 3.12 g = 60.58 g.
A heated metal sample is placed in water in an insulated calorimeter of negligible heat capacity. The table shows the data. What is the temperature of the metal and the water when they reach thermal equilibrium?
Answer and reasoning
A57.5°C A student who thinks the final temperature is the average of the two starting temperatures picks this. That holds only when m × c is the same for both; here the water's m × c (209 J/°C) is much larger than the metal's (45.0 J/°C).
B33.3°CCorrect Heat released by the metal equals heat absorbed by the water: (100.0)(0.450)(95.0 − T) = (50.0)(4.18)(T − 20.0), so 45.0(95.0 − T) = 209(T − 20.0) and T = 33.3°C. The water's much larger m × c keeps the final temperature close to the water's starting temperature.
C27.3°C A student who leaves out the masses, as if the heat for a temperature change did not depend on the amount, picks this: (0.450 × 95.0 + 4.18 × 20.0)/(0.450 + 4.18) = 27.3°C. The heat each sample exchanges is proportional to its mass, so the masses must be included.
D91.2°C A student who thinks a larger specific heat capacity gives a larger temperature change picks this, weighting each sample by m/c so that the water's temperature changes most. A larger c means a smaller temperature change for the same heat, so the water's temperature changes least.
Working Heat released by metal = heat absorbed by water: (100.0)(0.450)(95.0 − T) = (50.0)(4.18)(T − 20.0). 45.0(95.0 − T) = 209(T − 20.0); 4275 + 4180 = 254T; T = 33.3°C.
A 50.0 g block of metal at 90.0°C is placed in 100.0 g of water at 20.0°C in an insulated calorimeter of negligible heat capacity. At thermal equilibrium both are at 25.0°C. A student claims that the metal lost more heat than the water gained. Which statement best evaluates the claim?
Answer and reasoning
AThe claim is incorrect; no energy leaves the calorimeter, so the heat the metal lost equals the heat the water gainedCorrect Energy is conserved. No energy leaves an insulated calorimeter of negligible heat capacity, so qmetal + qwater = 0: the heat lost by the metal equals the heat gained by the water. The metal's temperature changes more only because its m × c is smaller.
BThe claim is correct; the metal's temperature fell by 65.0°C and the water's rose by only 5.0°C, so the metal lost more heat A student who treats heat and temperature as the same thing picks this. The size of a temperature change also depends on mass and specific heat capacity; the two heats are equal even though the temperature changes are not.
CThe claim is correct; some energy is used up as it passes from the metal to the water, so the water gains less heat A student who thinks energy is used up as it is transferred picks this. Energy is conserved: with no energy leaving the insulated calorimeter, all the energy the metal loses is gained by the water.
DThe claim is incorrect; the water gained more heat than the metal lost, as water has the larger specific heat capacity A student who thinks a substance with a larger specific heat capacity absorbs more heat picks this. The water cannot gain more energy than the metal releases; its larger specific heat capacity only means its temperature rises less for that heat.
An 80.0 g sample of a metal with a specific heat capacity of 0.450 J/(g·°C) cools from 85.0°C to 25.0°C. What is q for the metal?
Answer and reasoning
A+2.16 × 10³ J A student who thinks q is always positive picks this. The metal releases heat as it cools, so its energy decreases and q is negative.
B−2.16 × 10³ JCorrect ΔT = Tfinal − Tinitial = 25.0°C − 85.0°C = −60.0°C, so q = (80.0 g)(0.450 J/(g·°C))(−60.0°C) = −2.16 × 10³ J. The negative sign shows that the metal's energy decreases as it cools.
C−2.70 × 10¹ J A student who leaves out the mass, as if the heat for a temperature change did not depend on the amount, picks this: (0.450)(−60.0) = −27.0 J, the heat for just 1 g of the metal.
D+7.67 × 10³ J A student who adds 273 to the temperature change picks this, using ΔT = −60.0 + 273 = 213. A temperature change has the same value in °C and in K, so ΔT = −60.0 and q is negative.
Working ΔT = 25.0°C − 85.0°C = −60.0°C. q = mcΔT = (80.0 g)(0.450 J/(g·°C))(−60.0°C) = −2.16 × 10³ J; negative because the metal's energy decreases as it cools.
A student adds a solid to water in a foam-cup calorimeter and stirs. Before mixing, the solid and the water were both at 22.0°C. The graph shows the temperature of the mixture over time. Which conclusion about the dissolution process is supported by the data?
Answer and reasoning
AEndothermic, because energy is transferred from the water to the dissolving solidCorrect After the solid was added, the temperature of the mixture fell from 22.0°C to about 16°C. The water (the surroundings) lost thermal energy to the dissolving solid (the system), so the dissolution absorbed energy: it is endothermic.
BExothermic, because energy is transferred from the dissolving solid to the water A student who thinks the thermometer shows the energy of the dissolving solid itself picks this, reading the fall in temperature as energy released by the dissolution. The thermometer measures the water, which lost energy, so the dissolving solid absorbed it.
CExothermic, because energy is transferred from the water into the dissolving solid A student who has the direction of energy flow right but swaps the two terms picks this. A process that absorbs energy from its surroundings is endothermic.
DEndothermic, because cold is transferred from the dissolving solid to the water A student who thinks cold flows from a cold object into a warmer one picks this. No cold is transferred: the water cooled because it transferred energy to the dissolving solid, and that energy flow into the system is what makes the process endothermic.
A 60.0 g piece of hot metal is added to 100.0 g of water in a foam-cup calorimeter, and the temperature of the water is recorded as shown in the graph. The specific heat capacity of water is 4.18 J/(g·°C). Based on the graph, how much heat did the water absorb from the metal?
Answer and reasoning
A3.3 × 10³ JCorrect The graph shows the water starting at 22.0°C and reaching a maximum of 30.0°C, so ΔT = 8.0°C and q = (100.0 g)(4.18 J/(g·°C))(8.0°C) = 3.3 × 10³ J. The slow fall afterward shows the mixture losing energy to the room.
B1.3 × 10⁴ J A student who uses a temperature reading instead of the temperature change picks this, using 30.0°C as ΔT. The water started at 22.0°C, so ΔT = 8.0°C.
C2.5 × 10³ J A student who takes the last reading as the final temperature picks this, using ΔT = 28.0°C − 22.0°C = 6.0°C. After the maximum, the mixture is losing energy to the room, so the maximum, 30.0°C, is the better estimate of the temperature the water reached by absorbing heat from the metal.
D2.0 × 10³ J A student who uses the metal's mass in q = mcΔT picks this, (60.0 g)(4.18)(8.0). The temperature change and specific heat capacity are the water's, so the water's mass, 100.0 g, is needed.
Working From the graph: initial temperature 22.0°C; maximum 30.0°C; ΔT = 8.0°C. q = mcΔT = (100.0 g)(4.18 J/(g·°C))(8.0°C) = 3.3 × 10³ J. The slow fall after 120 s is energy lost to the room.
A student measures the mass of a metal sample, the heat it absorbs, and its temperature change, and calculates the metal's specific heat capacity in J/(g·°C). Which additional quantity does the student need in order to determine the metal's molar heat capacity in J/(mol·°C)?
Answer and reasoning
AThe specific heat capacity of liquid water A student who thinks water's specific heat capacity is needed in every heat calculation picks this. No water is involved in converting the metal's specific heat capacity into a molar heat capacity.
BThe sample's volume, to divide by 22.4 L/mol A student who thinks any substance occupies 22.4 L per mole picks this. That value applies only to an ideal gas at STP; the amount of a solid metal is found from its mass and molar mass.
CThe molar mass of the metal that makes up the sampleCorrect Molar heat capacity = specific heat capacity × molar mass, since J/(g·°C) × g/mol = J/(mol·°C). The student already has the specific heat capacity, so only the metal's molar mass is needed.
DNothing more, as the two heat capacities are equal A student who thinks specific and molar heat capacities are the same quantity picks this. One is per gram and the other is per mole, and they differ by a factor equal to the molar mass.
Working Cmolar = c × molar mass: J/(g·°C) × g/mol = J/(mol·°C). The only missing quantity is the molar mass of the metal.
In trial 1, a student dissolves 5.00 g of a solid in 50.0 g of water in a foam-cup calorimeter and records the temperature rise. In trial 2, the student repeats the procedure with 5.00 g of the same solid in 100.0 g of water at the same starting temperature. How do the temperature rise and the calculated heat released in trial 2 compare with those in trial 1?
Answer and reasoning
AThe temperature rise is about the same, and the heat released is twice as large A student who thinks a given amount of solid always produces the same temperature change picks this, then calculates q = mcΔT with about twice the mass. The same energy spread over about twice the mass gives about half the temperature rise.
BThe temperature rise is about twice as large, and the heat released is four times as large A student who thinks a larger amount of water draws more energy from the dissolving solid picks this. The energy released depends on the amount of solid, which is unchanged, so more water means a smaller temperature rise.
CThe temperature rise is about half as large, and the heat released is half as large A student who uses the mass of the solid in q = mcΔT picks this: with 5.00 g in both trials the calculated q follows ΔT, so half the temperature rise gives half the heat. The calculation must use the mass of the solution, which nearly doubles.
DThe temperature rise is about half as large, and the heat released is about the sameCorrect The same 5.00 g of solid dissolves, so about the same energy is released. That energy now warms about 105 g of solution instead of 55 g, so ΔT = q/(mc) is about half as large, and the calculation with the larger mass and smaller ΔT gives about the same heat.
A 50.0 g sample of metal X absorbs a certain amount of energy, and its temperature rises by 24.0°C. A 100.0 g sample of metal Y, whose specific heat capacity is half that of metal X, absorbs the same amount of energy. By how much does the temperature of metal Y rise?
Answer and reasoning
A48.0°C A student who leaves out the mass, as if the heat for a temperature change did not depend on the amount, picks this: halving c doubles ΔT to 48.0°C. Y's mass is also doubled, which halves ΔT again.
B24.0°CCorrect ΔT = q/(mc). For Y the heat is the same, the mass is doubled and the specific heat capacity is halved, so the product mc is unchanged and the temperature rise is the same, 24.0°C.
C12.0°C A student who thinks only the mass matters, not which substance it is, picks this: doubling the mass halves ΔT to 12.0°C. Y's specific heat capacity is half of X's, which doubles ΔT again.
D6.00°C A student who thinks a larger specific heat capacity gives a larger temperature change picks this, so that halving c halves ΔT and doubling m halves it again. A smaller c gives a larger temperature change for the same heat.
Working ΔT = q/(mc). For Y: q is the same, m is 2 times as large and c is ½ as large, so mc is unchanged (2 × ½ = 1). ΔTY = 24.0°C.
Equal masses of two liquids, X and Y, are heated, and neither liquid boils. The graph shows the temperature of each liquid as a function of the energy it has absorbed. The specific heat capacity of liquid X is 2.4 J/(g·°C). Based on the graph, what is the specific heat capacity of liquid Y?
Answer and reasoning
A1.6 J/(g·°C) A student who thinks the temperature change is proportional to the specific heat capacity picks this, scaling 2.4 J/(g·°C) by 20/30 because Y's temperature rises less. For the same heat and mass, the liquid with the smaller temperature change has the larger specific heat capacity.
B3.0 J/(g·°C) A student who uses the temperatures reached, 50°C and 40°C, in place of the temperature changes picks this: 2.4 × (50/40). Both liquids started at 20°C, so the changes are 30°C and 20°C.
C3.6 J/(g·°C)Correct For the same energy absorbed, 6 kJ, X warms by 30°C and Y by 20°C. With equal masses and equal q in q = mcΔT, c is inversely proportional to ΔT, so c of Y = 2.4 J/(g·°C) × (30/20) = 3.6 J/(g·°C).
D2.5 J/(g·°C) A student who adds 273 to each temperature change to convert it to kelvins picks this: 2.4 × (303/293). A change of 30°C is a change of 30 K, so the ratio of the changes is 30/20.
Working From the graph, after each liquid absorbs 6 kJ, X has warmed from 20°C to 50°C (ΔT = 30°C) and Y from 20°C to 40°C (ΔT = 20°C). For equal masses and equal q, c is inversely proportional to ΔT: cY = cX × (ΔTX/ΔTY) = 2.4 J/(g·°C) × (30/20) = 3.6 J/(g·°C).
Compiled from the AP Chemistry Course and Exam Description (effective Fall 2024) and our question bank · Specialist review in progress. How these pages are made · Free, no account
