Causes f Change Calrimetry Hw Des Energy Affect Change? Heat vs. Temerature HEAT TEMPERATURE Definitin: Deends n: Examles: Heat is energy and is measured in jules (J) r kiljules (kj). The symbl fr heat is H. Hw des heat energy travel? Frm t, NEVER frm t. Examles: We always talk in terms f HEAT, nt COLD. Heat f Physical Changes Recall that ne examle f a hysical change is when a substance ges thrugh a hase change (slid t liquid, liquid t slid, liquid t gas, etc.) LEAST Energy Middle Energy MOST ENERGY Least Heat Middle Heat Mst Heat Giving Off Heat Absrbing Heat We can calculate hw much heat a substance absrbs r gives ff. We just need 3 categries f infrmatin: 1. 2. 3.
What haened t substance? Heat is Sign f H..It biled..tem. change frm 23 C t 38 C..It cndensed..it melted..tem. changed frm 400 C t 130 C..It frze What haens t the temerature when a substance is changing hase (biling, melting, etc.)? Categries 1 and 2 (at bttm f revius age) are easy t handle. Abut Categry 3 Since all materials are unique, all materials have unique numbers that relate t heat energy. Fr examle, ure water bils at 100 C and melts at 0 C. Other substances change hases at ther secific temeratures. Examles: When a substance changes frm a slid t a liquid, a secific amunt f energy er gram f substance must be absrbed by the substance frm its surrundings WATER This same amunt f energy is released when the substance changes frm a liquid int a slid. C f = 333 J/g fr H 2 O We call this value, which is cnstant fr any articular substance, the heat f fusin. C v = 2256 J/g fr H 2 O
What des the heat energy absrbed r released during a hase change deend n? 1. 2. The frmula fr calculating heat energy gained r lst during a hase change is: H mc f ( fr slid liquid hase changes) H mcv ( fr liquid -gas hase changes ) Examle 1: A 65 g samle f ice at 0 C melts t becme water at 0 C. A. Des the samle absrb r release energy? B. Will the sign fr energy be (+) r (-)? C. Hw many jules (J) f energy are needed? + 21, 645 J Examle 2: A 154 g samle f water at 0 C freezes int ice at 0 C. A. Des the samle absrb r release energy? B. Will the sign fr energy be (+) r (-)? C. What is the change in temerature? D. Hw many jules (J) f energy are needed? - 51,282 J Within a hase, hwever, what haens when the substance ABSORBS energy? RELEASES energy? Fr finding the heat energy absrbed r released within a hase (i.e., with a change), what des the amunt f heat deend n? 1. 2. 3. The frmula fr calculating heat within a hase is: H m C T ( where ΔT change in temeratur e Tf Ti )
We call the cnstant C, the secific heat. Fr examle: C (ice) 2.077 J/(g C) C (water) 4.184 J/(g C) C (steam) 1.87 J/(g C) Examle 3: A 35 g samle f ice fes frm a temerature f 5 C t 23 C. A) Des the samle absrb r release energy? B) Will the sign fr energy be (+) r (-)? C) Hw many jules (J) f energy are needed? Examle 4: A 46 g samle f water ges frm a temerature f 34 C t 78 C. - 1309 J A) Des the samle absrb r release energy? B) Will the sign fr energy be (+) r (-)? C) Hw many jules (J) f energy are needed? We can visually sum u the energy changes exerience by a substance using a diagram. + 8460 J
Examle 5: A 80 g samle f ice at 10 C turns int water at 55 C. A) Des the samle absrb r release energy? B) Will the sign fr energy be (+) r (-)? C) Hw many jules (J) f energy are needed? 46,694 J Examle 6: A 34 g samle f steam at 120 C turns int water at 4 C. A) Des the samle absrb r release energy? B) Will the sign fr energy be (+) r (-)? C) Hw many jules (J) f energy are needed? -91,619 J Examle 7: In an insulated cntainer, 50 g f water (initially at 45 C are mixed with 90 g f water (initially at 82 C). What is the final temerature f the system? 68.8 C
Causes f Change Calrimetry KEY Hw Des Energy Affect Change? Heat vs. Temerature HEAT TEMPERATURE Definitin: Ttal kinetic energy f articles in a substance Average kinetic energy f articles In a substance Deends n: 1) articles average kinetic energy 2) number f articles (mass) 1) articles average kinetic energy Examles: Large beaker f warm water and small beaker have same temerature, but larger samle has Mre heat it can melt mre ice cubes. Heat is energy and is measured in jules (J) r kiljules (kj). The symbl fr heat is H. Hw des heat energy travel? Frm HOT t COLD, NEVER frm cld t ht. Examles: 1. Yu jum int cld swimming l, yur bdy heat leaves and yu feel cld. ( cld frm water des nt g int yu.) 2. Ta water, thrw in ice cubes, heat frm water ges int cld cubes; cld frm cubes des nt g int warm water. We always talk in terms f HEAT, nt COLD. 50 g Ice Cube 50 g Water 50 g Steam Less Heat Mre Heat Even Mre Heat Heat f Physical Changes Recall that ne examle f a hysical change is when a substance ges thrugh a hase change (slid t liquid, liquid t slid, liquid t gas, etc.) LEAST Energy Middle Energy MOST ENERGY SOLID LIQUID GAS Least Heat Middle Heat Mst Heat (slid liquid) MELTING FREEZING (liquid slid) (liquid gas) BOILING CONDENSING (gas liquid) Giving Off Heat Absrbing Heat We can calculate hw much heat H a substance absrbs (+) H r gives ff (-) H. We just need 3 categries f infrmatin: 1. MASS f substance m 2. beginning (T i ) and final (T f ) TEMPERATURE T = T f - T i 3. CONSTANTS assciated with the substance C = secific heat
What haened t substance? Heat is Sign f H..It biled heat + H 2 O(l) H 2 O(g) Added (+)..Tem. change frm 23 C t 38 C Added (+)..It cndensed H 2 O(g) H 2 O(l) + heat Remved (-)..It melted heat + slid liquid Added (+)..Tem. changed frm 400 C t 130 C Remved (-)..It frze liquid slid + heat Remved (+) What haens t the temerature when a substance is changing hase (biling, melting, etc.)? It stays the same. Categries 1 and 2 (at bttm f revius age) are easy t handle. Abut Categry 3 Since all materials are unique, all materials have unique numbers that relate t heat energy. Fr examle, ure water bils and cndenses at 100 C and melts and freezes at 0 C. Other substances change hases at ther secific temeratures. Examles: Irn melts (and freezes) at ~1520 C Gld melts (and freezes) at ~1040 C Helium bils (and cndenses) at ~ -268 C Nen bils (and cndenses) at.. ~ -246 C When a substance changes frm a slid t a liquid, a secific amunt f energy er gram f substance must be absrbed by the substance frm its surrundings Heat frm envirnment is absrbed by ice. HEAT Heat frm water is released int freezer. WATER ice at 0 C water at 0 C HEAT This same amunt f energy is released when the substance changes frm a liquid int a slid. (liquid slid) C f = 333 J/g fr H 2 O C f = heat f fusin We call this value, which is cnstant fr any articular substance, the heat f varizatin. (liquid gas) C v = 2256 J/g fr H 2 O
What des the heat energy absrbed r released during a hase change deend n? 1. heat f fusin, C f (r heat f varizatin, C v ) 2. mass f samle The frmula fr calculating heat energy gained r lst during a hase change is: H mc f ( fr slid liquid hase changes) H mcv ( fr liquid -gas hase changes ) Examle 1: A 65 g samle f ice at 0 C melts t becme water at 0 C. A. Des the samle absrb r release energy? B. Will the sign fr energy be (+) r (-)? C. Hw many jules (J) f energy are needed? x J H mc f H (65 g)(333 J g) + 21, 645 J Examle 2: A 154 g samle f water at 0 C freezes int ice at 0 C. A. Des the samle absrb r release energy? B. Will the sign fr energy be (+) r (-)? C. What is the change in temerature? There is NO change in temerature. D. Hw many jules (J) f energy are needed? x J H mc f H (154 g)(333 J g) Within a hase, hwever, what haens when the substance - 51,282 J ABSORBS energy? Its temerature increases RELEASES energy? Its temerature decreases Fr finding the heat energy absrbed r released within a hase (i.e., with a temerature change), what des the amunt f heat deend n? 1. hw big the temerature change is 2. hw big the samle is (mass) 3. materials cnstants The frmula fr calculating heat within a hase is: H m C T ( where ΔT change in temeratur e Tf Ti )
We call the cnstant C, the secific heat. Fr examle: C (ice) 2.077 J/(g C) C (water) 4.184 J/(g C) C (steam) 1.87 J/(g C) Examle 3: A 35 g samle f ice ges frm a temerature f 5 C t 23 C. a. Des the samle absrb r release energy? b. Will the sign fr energy be (+) r (-)? c. Hw many jules (J) f energy are needed? H = m C T (where T = change in temerature = T T ) f f i J (35 g)(2.077 J g C)( 23 C ( 5 C)) Examle 4: A 46 g samle f water ges frm a temerature f 34 C t 78 C. - 1309 J D) Des the samle absrb r release energy? E) Will the sign fr energy be (+) r (-)? F) Hw many jules (J) f energy are needed? x J m C T x J = (46 g)(4.184 J g C C C )(78 34 ) x = 8460 J We can visually sum u the energy changes exerience by a substance using a diagram. + 8460 J
Examle 5: A 80 g samle f ice at 10 C turns int water at 55 C. D) Des the samle absrb r release energy? E) Will the sign fr energy be (+) r (-)? F) Hw many jules (J) f energy are needed? ice warms u t 0 C H = m C T = (80 g)(2.077 J g C )(0 10 ) 1661.6 J ice turns t water at 0 C H = m C f = (80 g)(333 J/g) = + 26,640 J )(55 0 ) water warms u t 55 C H = m C T = (80 g)(4.1840 J g C = + 18,392 J 46,694 J Examle 6: A 34 g samle f steam at 120 C turns int water at 4 C. D) Des the samle absrb r release energy? E) Will the sign fr energy be (+) r (-)? F) Hw many jules (J) f energy are needed? steam cls t 100 C H = m C T = (34 g)(1.87 J g C)(100 120 C ) = - 1,271.6 J steam turns t water at 100 C H = m C v = (34 g)(2256 J g ) = - 76,704 J water cls frm 100 t 4 C H = m C T = (34 g)(4.184 J g C)(4 C - 100 C) = - 13,643.5 J -91,619 J Examle 7: In an insulated cntainer, 50 g f water (initially at 45 C are mixed with 90 g f water (initially at 82 C). What is the final temerature f the system? X final temerature H1 m C T = (50 g)(4.184 J g C)( X 45 C ) gains energy (+) H1 m C T = (90 g)(4.184 J g C)( X 82 C ) lses energy (-) H - H 1 2 (50 g)(4.184 J g C X C = - (90 g)(4.184 J g C X C )( 45 ) )( 82 ) 68.8 C