Or the amount of energy needed to heat substances up Specific Heat Capacity Or the amount of energy needed to heat substances up

When I eat hot pizza or a melted cheese sandwich, the cheese feels a lot hotter than the crust or bread: in particular, the cheese might scald the roof of my mouth, but the crust will not. Why … Why … Why???

Have you ever noticed that on a hot summer day that the ocean is cooler than the hot sand? Why? The sun has been beating down on (delivering energy to) both of them for the same amount of time.....same amount of energy.

Water absorbs a lot of heat energy before its temperature changes while sand needs little heat energy before its temperature increases.

How does the number of particles influence thermal energy? So… we know that Temperature is related to the average kinetic energy of the particles in a substance. How does the number of particles influence thermal energy?

Two towels – same size/mass You can add the same amount of water (heat), but the cheap towel will be “wetter” You can add the same amount of water (heat), but the cheaper towel will be “wetter” (temperature). They have different capacities for absorbing water

The flow of thermal energy from one object to another. Cup gets cooler while hand gets warmer Heat The flow of thermal energy from one object to another. Heat always flows from warmer to cooler objects. Ice gets warmer while hand gets cooler

The heat transferred is proportional to the mass of the object, the specific heat capacity of the object and the temperature change the object undergoes. Heat = Q (J) Mass = m (kg) Specific heat = c (J/kg˚C) Temperature change = ∆T ˚C (Tf – Ti)

Q = mcDT mass specific heat capacity Quantity of heat temperature change

Specific heat capacity (c) Specific heat capacity is the amount of energy needed to raise the temperature of 1 kg of a substance by 1°C

= Q DT m c Q m c ∆T = Q m cDT = Q c mDT

Things heat up or cool down at different rates based on specific heat capacity. Land heats up and cools down faster than water, and aren’t we lucky for that!?

C water = 4184 J / kg C (“holds” its heat) Specific heat is the amount of heat required to raise the temperature of 1 kg (but in Chem we use g) of a material by one degree (C or K, they’re the same size). C water = 4184 J / kg C (“holds” its heat) C sand = 664 J / kg C (less Q to change) This is why land heats up quickly during the day and cools quickly at night and why water takes longer.

Questions

Heat Transfer A 32-g silver spoon cools from 60°C to 20°C. How much heat is lost by the spoon?? GIVEN: m = 32 g Ti = 60°C Tf = 20°C c = 235 J/kg·C Q = ? WORK: Q = m·c·T m = 32 g = 0.032 kg 1kg = 1000g T = 20°C - 60°C = – 40°C Q = (0.032kg)(235J/kg·C)(-40°C) Q = – 301 J

Heat Transfer How much heat is required to warm 230 g of water from 12°C to 90°C? GIVEN: m = 230 g Ti = 12°C Tf = 90°C c= 4184 J/kg·C Q = ? WORK: Q = m·c·T m = 230 g = 0.23 kg T = 90°C - 12°C = 78°C Q = (0.23kg)(4184 J/kg·C)(78°C) Q = 75,061 J

Q = m c DT Q = (0.0256kg)(4184J/kgC)(30.0C) Q = 3213 J How much heat is needed to raise the temperature of 25.6 grams of water from 20.0 C to 50.0 C? Q = m c DT Q = (0.0256kg)(4184J/kgC)(30.0C) Q = 3213 J

For example 0.5 kg of olive oil is heated until its temperature rises by 120 °C. If the specific heat capacity of olive oil is 1970 J/kg/°C, how much heat energy was used? Q = Mass x Specific Heat capacity x Temp rise Q = 0.5 x 1970 x 120 Q = 118200 J

q = - 6136 J Specific Heat Capacity If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al? heat gain/lose = q = (c)(mass)(∆T) where ∆T = Tfinal - Tinitial q = (0.025 g)(899 J/gC)(37 - 310)C q = - 6136 J Notice that the negative sign on q signals heat “lost by” or transferred OUT of Al.

q transferred = (c)(mass)(∆T) Heat can be Transferred even if there is No Change in State q transferred = (c)(mass)(∆T)

Or… Heat Transfer can cause a Change of State Changes of state involve energy (at constant T) Ice + 333 J/g (heat of fusion) -----> Liquid water Is there an equation? Of course! q = (heat of fusion)(mass)

Heat Transfer and Changes of State Liquid (l)  Vapor (g) Requires energy (heat). Why do you… cool down after swimming ??? use water to put out a fire???

Remember this – it’s the Heating/Cooling Curve for Water! Woa! Evaporate water Note that T is constant as a phase changes

q transferred = (c)(mass)(∆T) So, let’s look at this equation again… q = (heat of fusion)(mass) (There’s also q = (heat of vaporization)(mass), by the way, for when we are talking about vaporization) WHY DO I NEED THIS WHEN I HAVE q transferred = (c)(mass)(∆T) HUH??? Well, when a phase changes THERE IS NO change in temperature… but there is definitely a change in energy!

So… if I want the total heat to take ice and turn it to steam I need 3 steps… 1) To melt the ice I need to multiply the heat of fusion with the mass…q = (heat of fusion)(mass)

2) Then, there is moving the temperature from 0 C to 100C… for this there is a change in temperature so we can use… q transferred = (c)(mass)(∆T) 3) But wait, that just takes us to 100 C, what about vaporizing the molecules? Well, then we need q = (heat of vaporization)(mass)…

Add ‘em all up and there it is! Now, lucky for us, just like there are tables for specific heats, there are also tables for heats of fusion and heats of vaporization. Whew, At least we don’t have to worry about that!

Heat of fusion of ice = 333 J/g Specific heat of water = 4.2 J/g•K Heat & Changes of State What quantity of heat is required to melt 500. g of ice and heat the water to steam at 100 oC? Heat of fusion of ice = 333 J/g Specific heat of water = 4.2 J/g•K Heat of vaporization = 2260 J/g +2260 J/g +333 J/g

And now… More! Heat & Changes of State How much heat is required to melt 500. g of ice and heat the water to steam at 100 oC? 1. To melt ice q = (500. g)(333 J/g) = 1.67 x 105 J 2. To raise water from 0 oC to 100 oC q = (500. g)(4.2 J/g•K)(100 - 0)K = 2.1 x 105 J 3. To evaporate water at 100 oC q = (500. g)(2260 J/g) = 1.13 x 106 J 4. Total heat energy = 1.51 x 106 J = 1510 kJ

Aka… How we Measure Heats of Reaction CALORIMETRY Aka… How we Measure Heats of Reaction In a Constant Volume or “Bomb” Calorimeter, we burn a combustible sample. From the heat change , we measure heat evolved in a reaction to get ∆E for reaction.

Total heat evolved = qtotal = qwater + qbomb BOOM! Combustible material ignited at constant volume! This heats up the “bomb”, which heats up the water surrounding it… First, some heat from reaction warms the water, which we know the mass of and “c” for… qwater = (c)(water mass)(∆T) THEN, some heat from reaction warms “bomb,” which has a known specific heat for the entire apparatus (typically), so we don’t need the mass… qbomb = (heat capacity, J/K)(∆T) Total heat evolved = qtotal = qwater + qbomb

Calculate heat of combustion of octane. The Mathy bit… Measuring Heats of Reaction using… CALORIMETRY Calculate heat of combustion of octane. C8H18 + 25/2 O2 --> 8 CO2 + 9 H2O You could burn 1.00 g of octane… or you could just note that… Temp rises from 25.00 to 33.20 oC Calorimeter contains 1200 g water Heat capacity of bomb = 837 J/K

VIOLA! Heat of combustion of 1.00 g of octane = - 48.0 kJ Step 1 Calc. heat transferred from reaction to water. q = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 J Step 2 Calc. heat transferred from reaction to bomb. q = (bomb heat capacity)(∆T) = (837 J/K)(8.20 K) = 6860 J Step 3 Total heat evolved 41,170 J + 6860 J = 48,030 J Heat of combustion of 1.00 g of octane = - 48.0 kJ

H = (5,650 grams H2O)(4.184 J/g0C)(55.40 C) One More (ok, LAST) Example… With a twist… If I burn 0.315 moles of hexane (C6H14) in a bomb calorimeter containing 5.65 liters of water, what’s the molar heat of combustion of hexane is the water temperature rises 55.40 C? The heat capacity of water is 4.184 J/g0C. H = mCpT H = (5,650 grams H2O)(4.184 J/g0C)(55.40 C) H = 1310 kJ Do you think we’re done?

NOPE! We were asked for the MOLAR heat of combustion! Tricky, tricky… The amount of energy released in burning completely one mole of substance. Vs. Heat of combustion The amount of heat released per unit mass or unit volume of a substance when the substance is completely burned. We also have heat of fusion vs MOLAR heat of fusion and heat of vaporization vs MOLAR heat of vaporization, so be watchful…

What we calculated is the amount of energy generated when 0 What we calculated is the amount of energy generated when 0.315 moles of hexane is burned, which is close but not quite what we were asked for... To find the molar heat of combustion, we need to multiply this by (1 mole/0.315 moles) = 3.17. As a result, the molar heat of combustion of hexane is 4150 kJ/mol.