Arrow Lab

Química Geral · Thermochemistry

Calorimetria e capacidade calorífica

Uma lição resolvida gratuita e 40 problemas de prática em que você mesmo desenha cada seta curva.

Travou em calorimetria e capacidade calorífica? Escaneie o problema.

Tire uma foto da reação da sua lista ou do livro e desenhe o mecanismo com cada seta corrigida. 1 scan grátis agora, sem cadastro; mais 1 com uma conta gratuita.

Escaneie sua lição de graça

O que você vai aprender

  • Distinguish heat capacity, specific heat capacity and molar heat capacity, and convert between them with mass and molar mass.
  • Calculate heat, temperature change, mass or specific heat with q = mcΔT, using the sign convention that heat absorbed is positive.
  • Turn coffee-cup calorimetry data into molar enthalpies of dissolution and neutralization, including limiting reagents and the heat capacity of the cup.
  • Calibrate a bomb calorimeter with benzoic acid, use its calorimeter constant to find energies of combustion, and relate ΔU to ΔH.
  • Find the final temperature when bodies exchange heat, including melting ice and condensing steam.

Lição gratuita

O texto da lição resolvida. No Arrow Lab, cada etapa vem com as estruturas e as setas desenhadas.

Ler a aula gratuita (7 partes)

Three ways to state how much heat a temperature change needs

Heat capacity C is the heat needed to raise the temperature of a whole object by 1 °C, in J/°C. It is extensive: twice the sample has twice the heat capacity. Specific heat capacity c is the heat capacity per gram, in J/(g·°C), and molar heat capacity Cm is the heat capacity per mole, in J/(mol·°C); both are intensive properties of the substance. They are linked by C = m × c and Cm = c × M, where M is the molar mass. Liquid water (drawn) has the unusually large specific heat 4.184 J/(g·°C), so it warms and cools slowly. Copper has c = 0.385 J/(g·°C), so Cm = 0.385 × 63.55 = 24.5 J/(mol·°C). Most solid metals have molar heat capacities near 25 J/(mol·°C), which is why a heavy metal such as gold (0.129 J/(g·°C)) has a small specific heat: a gram of it contains few atoms. A temperature difference has the same size in kelvin and in degrees Celsius, so J/(g·K) and J/(g·°C) are the same unit.

q = mcΔT and the sign of q

The heat absorbed by a sample that changes temperature without changing phase is q = m × c × ΔT, with ΔT = T(final) − T(initial). Heat absorbed is positive and heat released is negative. Warming 500.0 g of water from 22.0 °C to 32.0 °C needs q = 500.0 g × 4.184 J/(g·°C) × 10.0 °C = 20 920 J = 20.9 kJ. For an object of known heat capacity, q = C × ΔT; for several parts warmed together, such as a pan and the water in it, add their heat capacities. In calorimetry the reaction is the system and the water, solution and container are the surroundings. Energy is conserved, so q(reaction) = −q(surroundings): if the solution warms, q(solution) > 0 and the reaction is exothermic, with q(reaction) < 0.

Coffee-cup calorimetry: constant pressure

A nested polystyrene cup open to the air works at constant pressure, so the heat of the process equals its enthalpy change: q(p) = ΔH. Measure the temperature change of the solution, calculate q(solution) = m c ΔT with the total mass of the solution (water plus dissolved solid, or both mixed solutions), add C(cup) × ΔT if the cup's heat capacity is known, and set q(reaction) = −q(solution + cup). Divide by the moles that reacted to get ΔH in kJ/mol. Example: 4.00 g of NaOH (0.100 mol, drawn) dissolves in 96.0 g of water and the 100.0 g of solution warms by 10.6 °C. q(solution) = 100.0 × 4.184 × 10.6 = 4435 J, so ΔH(soln) = −4435 J ÷ 0.100 mol = −44.4 kJ/mol. A solution that cools, as with ammonium nitrate, signals an endothermic process with ΔH > 0. For neutralization, the moles of water formed come from the limiting reagent. Every strong acid with every strong base has the same net reaction, H₃O⁺ + OH⁻ → 2 H₂O, and about −56 kJ per mole of water. Because a real cup slowly loses heat, the temperature–time line after mixing is extrapolated back to the moment of mixing to find the true ΔT.

Bomb calorimetry: constant volume

A weighed sample is burned in excess oxygen inside a sealed steel bomb sitting in a bucket of water. The volume is fixed, so no pressure–volume work is done and the heat measured is the change in internal energy: q(v) = ΔU. The whole assembly (bomb, water, stirrer, thermometer) is treated as one object with a calorimeter constant C(cal) in kJ/°C, so q(cal) = C(cal) × ΔT and q(reaction) = −q(cal). C(cal) is found by burning a standard of accurately known energy of combustion, usually benzoic acid (drawn), ΔcU = −26.43 kJ/g. Example: 1.000 g of benzoic acid raises the temperature by 2.59 °C, so C(cal) = 26.43 kJ ÷ 2.59 °C = 10.2 kJ/°C. The same water mass must be used in every later run, or the constant no longer applies. If C(cal) is split into parts, add them: C(cal) = C(hardware) + m(water) × 4.184 J/(g·°C). To convert to an enthalpy, ΔH = ΔU + Δn(gas)RT, where Δn(gas) is the moles of gaseous products minus gaseous reactants in the balanced equation; when Δn(gas) = 0, as for glucose, ΔcH = ΔcU.

Heat exchange to a common final temperature

When a hot body and a cold body touch in an insulated container, heat flows until they reach the same final temperature T(f). No heat is lost, so q(hot) + q(cold) = 0, or m₁c₁(T(f) − T₁) + m₂c₂(T(f) − T₂) = 0. Without phase changes this gives T(f) = (m₁c₁T₁ + m₂c₂T₂) ÷ (m₁c₁ + m₂c₂): a heat-capacity-weighted average, not a simple average. Mixing 100.0 g of water at 70.0 °C with 300.0 g at 10.0 °C gives T(f) = (7000 + 3000) ÷ 400.0 = 25.0 °C. The same balance run backwards gives an unknown specific heat or an unknown starting temperature: a metal sample dropped into water loses exactly the heat the water gains, so c(metal) = m(w)c(w)ΔT(w) ÷ (m(metal) × |ΔT(metal)|). A metal of small specific heat changes temperature a lot while the water barely moves.

Phase changes inside the calorimeter

Melting, boiling and condensing happen at constant temperature, and the heat they need or release is q = m × ΔH(phase change). For water, ΔHfus = 334 J/g (6.01 kJ/mol) at 0 °C and ΔHvap = 2260 J/g (40.7 kJ/mol) at 100 °C. Melting absorbs heat and freezing releases the same amount; vaporization absorbs heat and condensation releases it. Treat each stage separately and keep the signs straight. Example: 20.0 g of ice at 0 °C is added to 100.0 g of water at 50.0 °C. Melting all the ice needs 20.0 × 334 = 6680 J, and the water could release 100.0 × 4.184 × 50.0 = 20 900 J before reaching 0 °C, so all the ice melts. Then 6680 + 20.0 × 4.184 × T(f) = 100.0 × 4.184 × (50.0 − T(f)), so T(f) = 28.4 °C. Always make that check first: if the warm water cannot supply enough heat to melt all the ice, the mixture ends at 0 °C with some ice left, and the mass melted is the available heat ÷ 334 J/g. Steam is the reverse case: each gram releases 2260 J on condensing before the condensed water cools, which is why steam heats far more than the same mass of boiling water.

What to carry forward

Use C for an object, c per gram and Cm per mole, with q = mcΔT and positive q for heat absorbed. In any calorimeter, the heat gained by the surroundings is the heat lost by the system: ΔH = −q(solution + cup)/n at constant pressure, ΔU = −C(cal)ΔT/n at constant volume, then ΔH = ΔU + Δn(gas)RT. For bodies exchanging heat, set the sum of all q to zero, include q = mΔH for any melting or condensing, and check first whether a phase change goes to completion.

Problemas de prática (40)

Praticar é grátis com uma conta.

Preveja o produto 3

Coffee-cup calorimetry

  • Twice the volume of each solutionPredict how the temperature rise of a neutralization changes when both reagent volumes are doubled at fixed concentration.
  • Changing the strong acid and the strong basePredict the molar enthalpy of neutralization of a new strong acid–strong base pair from the net ionic equation.

Heat exchange and phase changes

  • A hot nail in cold waterPredict the conditions that hold when two bodies exchanging heat reach thermal equilibrium.

Classificar 4

Bomb calorimetry

  • What does a bomb calorimeter measure?Classify the heat measured at constant volume as the change in internal energy.
  • When does the bomb give ΔH directly?Classify combustions by whether their moles of gas change, and so whether ΔcH equals ΔcU.

Coffee-cup calorimetry

  • Ammonium nitrate dissolving in a coffee cupClassify a dissolution as endothermic or exothermic from the sign of the solution's temperature change.

Heat capacity and q = mcΔT

  • Which heat quantity depends on sample size?Classify heat capacity as extensive and specific and molar heat capacity as intensive.

Comparar 2

Heat exchange and phase changes

  • Rank the heat released by 1 g of waterRank heats released by sensible cooling and by phase changes of water using q = mcΔT and q = mΔH.

Heat capacity and q = mcΔT

  • Same heat, different temperature risesRank the temperature rise of equal masses given the same heat, from their specific heats.

Calcular 22

Coffee-cup calorimetry

  • Enthalpy of solution of ammonium nitrateCalculate a molar enthalpy of solution from the temperature drop of the whole solution in a coffee-cup calorimeter.
  • Enthalpy of solution of CaCl₂, counting the cupInclude the heat capacity of the calorimeter cup when calculating a molar enthalpy of solution.
  • Calibrating a coffee cup with hot and cold waterFind the heat capacity of a calorimeter cup from the heat imbalance when hot and cold water are mixed in it.
  • Temperature rise with a limiting basePredict the temperature rise of a neutralization in which the base is the limiting reagent.

Bomb calorimetry

  • Calibrating a bomb with benzoic acidCalculate a bomb calorimeter constant from the combustion of a weighed benzoic acid standard.
  • From ΔcU to ΔcH for benzoic acidConvert a bomb calorimeter ΔcU into ΔcH using the change in moles of gas in the balanced combustion equation.
  • Heat of a combustion from the calorimeter constantCalculate the heat of a reaction from a bomb calorimeter's constant and temperature rise, with the correct sign.
  • Glucose in a bomb with hardware and waterAssemble a bomb calorimeter's total heat capacity from its hardware and water, then find a molar energy of combustion.
  • Energy of combustion of naphthaleneConvert a bomb calorimeter temperature rise into a molar internal energy of combustion.
  • Food Calories in a peanutConvert a bomb calorimeter result into nutritional Calories per gram.

Heat exchange and phase changes

  • Ice from the freezer in warm waterCalculate a final temperature that requires warming ice, melting it and warming the meltwater.
  • Final temperature of copper in waterCalculate the final temperature reached by a hot metal and cool water from their masses and specific heats.
  • Too much ice to meltRecognize when warm water cannot melt all the ice and calculate the mass of ice left at 0 °C.
  • Ice cube in warm waterCalculate the final temperature when ice at 0 °C melts completely in warm water.
  • Specific heat of a metal by the method of mixturesFind the specific heat of a metal from the temperature changes when it is dropped into water.
  • Mixing hot and cold waterCalculate the final temperature of two water samples mixed in an insulated container.
  • Bubbling steam into cold waterCalculate the final temperature when steam condenses in cold water, counting both condensation and cooling of the condensate.

Heat capacity and q = mcΔT

  • Heat to warm a mug of waterCalculate the heat absorbed by water from its mass, specific heat and temperature change.
  • How long to heat a pan of water?Combine the heat capacities of a container and its contents to find the heating time at a given power.
  • Heating an iron skilletCalculate the heat absorbed by an object after converting its mass to the units of the specific heat.
  • Molar heat capacity of copperConvert a specific heat capacity into a molar heat capacity using the molar mass.
  • Specific heat of a metal sampleCalculate a specific heat capacity from the heat absorbed, the mass and the temperature change.

Encontre o erro 3

Bomb calorimetry

  • Less water than during calibrationDiagnose how using a calorimeter constant measured with a different water mass biases a combustion result.

Heat capacity and q = mcΔT

  • Find the error in a heat calculationDiagnose the error of adding 273.15 to a temperature difference.

Coffee-cup calorimetry

  • Why is the measured ΔH too small?Diagnose which experimental error makes a measured enthalpy of neutralization smaller in magnitude than the accepted value.

Interprete as evidências 1

  • Correcting a temperature–time curve for heat lossInterpret a calorimeter temperature–time record by extrapolating the cooling line back to the time of mixing.

Trabalhe de trás para frente 1

  • How hot was the copper?Work backwards from a measured final temperature to the starting temperature of a hot metal.

Planeje uma síntese 1

  • How to find a bomb calorimeter's constantChoose the procedure that determines a bomb calorimeter's heat capacity with a combustion standard.

Explique por quê 3

Coffee-cup calorimetry

  • Why a coffee cup gives ΔHExplain why the heat measured in an open coffee-cup calorimeter equals the enthalpy change.

Heat capacity and q = mcΔT

  • Why gold warms so easilyExplain differences in specific heat of metals by their nearly equal molar heat capacities and different molar masses.

Heat exchange and phase changes

  • Why steam burns worse than boiling waterExplain the severity of steam burns by the heat released on condensation.

Travou em calorimetria e capacidade calorífica? Escaneie o problema.

Tire uma foto da reação da sua lista ou do livro e desenhe o mecanismo com cada seta corrigida. 1 scan grátis agora, sem cadastro; mais 1 com uma conta gratuita.

Escaneie sua lição de graça