ENTHALPY CHANGES FOR A CHEMICAL REACTION scaling a rxn up or down (proportionality) quantity 1 from rxn heat 1 from Δ r H. = 32.

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1 CHEMISTRY 103 Help Sheet #10 Chapter 4 (Part II); Sections Do the topics appropriate for your lecture Prepared by Dr. Tony Jacob (Resource page) Nuggets: Enthalpy of a rxn (proportionality calculations); Hess s Law; Bomb Calorimetry; Coffee Cup Calorimetry; Δ f H ; Phase Change + Heat Transfer; ENTHALPY CHANGES FOR A CHEMICAL REACTION scaling a rxn up or down (proportionality) quantity 1 from rxn heat 1 from Δ r H = quantity 2 heat 2 quantity 1 = #mol from balanced rxn; heat 1 = heat from balanced rxn = Δ r H; quantity 2 = mass or mol given in the question; heat 2 = heat associated with quantity 2 Example 1: How much heat is released when 16.0g O 2 reacts in this rxn: 2CuS + 3O 2 2CuO + 2SO 2 Δ r H = -193 kj/mol 3molO Answer: 2 193kJ /mol = 16.0gO 2 3molO ; 2 x 193kJ /mol = 0.500molO 2 ; x = 0.500molO 2 ( 193kJ) = 32.2kJ x 3molO 2 Another way to set up this problem: 16gO2 1mol O 2 193kJ = 32.2kJ 32g O2 3mol O2 Example 2: When 12.5g Fe is reacted in the following reaction 19.9kJ are evolved. What is the enthalpy of reaction, Δ r H for the reaction? 3Fe(s) + 2O 2 (g) Fe 3 O 4 (s) 12.5g Fe 3mol Fe mol Fe 3mol Fe Answer: = ; = ; x = -59.7; x = kJ = Δ 19.9kJ x 19.9kJ x r H 19.9kJ 55.85g Fe Another way to set up this problem: 3mol Fe = 266.7kJ = Δ r H 12.5g Fe 1mol Fe HESS S LAW: If 2 or more rxns are added, the Δ r H of the net rxn is the sum of the Δ r H for each rxn added. 1. Reverse reaction: change the sign of ΔH Δ forward H = Δ reverse H 2. Multiply a reaction by a coefficient, c: multiply ΔH by c Δ new H = c(δ r H) 3. Add reactions: add ΔH s: Δ net-rxn H = Δ r1 H + Δ r2 H +... (Hess's Law) CALORIMETRY an experimental technique used to measure heat transfer and from this determine Δ r H. Constant Volume Calorimetry V is constant; a bomb calorimeter is often used to do these measurements Constant Pressure Calorimetry P is constant; a coffee cup calorimeter is often used to do these measurements Typical Calorimetry Questions I. Given Δ r H find T f I. Given T f find Δ r H Step 1: Find q sys mol Proportionality: reacted q sys = mol balanced rxn Δ r H Step 2: Find q surr q surr = -q sys Step 3: Find T f Coffee Cup: q surr = C soln m total (T f T i ) Bomb: q surr = q bomb calor + q soln q surr = C calorimeter const (T f T i ) + C soln m total (T f T i ) Step 1: Find q surr Coffee Cup: q surr = C soln m total (T f T i ) Bomb: q surr = q bomb calor + q soln q surr = C calorimeter const (T f T i ) + C soln m total (T f T i ) Step 2: Find q sys q sys = -q surr Step 3: Find Δ r H Proportionality: mol reacted q sys = mol balanced rxn Δ r H (Note: In bomb calorimeters you are calculating ΔE rather than Δ r H since volume is constant!)

2 BOMB CALORIMETRY Example: 4.50g of ethanol, CH 3 CH 2 OH, was burned in a constant volume bomb calorimeter (calorimeter constant = 1285J/ C) which contained 125g water. After the reaction, the temperature of the water and calorimeter increased from to C. What is the Δ r H for CH 3 CH 2 OH(l)? Start by writing the combustion reaction for CH 3 CH 2 OH(l). C soln = 4.184J/g C Answer: -1236kJ {Write rxn: CH 3 CH 2 OH(g) + 3O 2 (l) 2CO 2 (g) + 3H 2 O(g); Step1 (find q surr ): q surr = q calorimeter + q H2O ; q surr = C cal ΔT + C soln m mix ΔT; q surr = (1285J/ C)( C) + (4.184J/g C)( g)( C) = 84,926J + 35,809J = 120,735J; Step 2 (find q sys ): q sys = -q surr ; q sys = -120,735J; Step 3 (find Δ r H): Proportionality: 4.50g CH 3 CH 2 OH/(-120,735J) = 1mol CH 3 CH 2 OH/Δ r H ; convert 1 mol CH 3 CH 2 OH to grams: 1mol CH 3 CH 2 OH x (46.07g CH 3 CH 2 OH/1mol CH 3 CH 2 OH) = 46.07g CH 3 CH 2 OH; 4.50g CH 3 CH 2 OH/-120,735J = 46.07g CH 3 CH 2 OH/Δ r H ; 4.50Δ r H = -5,562,280J; Δ r H = -1,236,062J x 1kJ/1000J = -1236kJ} (Note: In bomb calorimeters you are actually calculating ΔE rather than Δ r H since volume is constant!) COFFEE CUP CALORIMETRY Example 1: When 2.50g H 2 SO 4 (l) at 21.5 C was added to 75.0g water at 21.5 C, the temperature of the solution increased to 29.0 C. What is the value of Δ r H for H 2 SO 4 (l) H 2 SO 4 (aq)? C soln = 4.184J/g C Answer: -95.4kJ {Step 1 (find q surr ): q surr = C soln m(t f T i ) = (4.184J/g C)( )( ) = 2432J; Step 2 (find q sys ): q sys = -q surr ; q sys = -2432J; Step 3 (find Δ r H): proportionality; 2.50g H 2 SO 4 /-2432J = 1mol H 2 SO 4 /Δ r H; 2.50g H 2 SO 4 /-2432J = 98.09g H 2 SO 4 /Δ r H; (2.5)(Δ r H) = (-2432)(98.09); Δ r H = J x 1kJ/1000J = -95.4kJ} Example 2: 10.0g KOH(s) is dissolved in 75.0ml water in a coffee cup calorimeter. The water and KOH are initially at 25.0 C. What is the final temperature? C soln = 4.184J/g C, D soln = 1.0g/ml. KOH(s) KOH(aq) Δ r H = -57.6kJ/mol Answer: 54.0 C {Step 1 (find q sys ): Proportionality: 10.0g KOH/(q sys ) = 1mol KOH/-57.6kJ; convert 10.0g KOH to mol: 10.0g KOH x (1mol KOH/55.1g KOH) = 0.178mol KOH; 0.178mol KOH/(q sys ) = 1mol KOH/-57.6kJ; q sys = -10.3kJ Step 2 (find q surr ): q surr = -q sys q surr = +10.3kJ Step 3 (find T f ): m water = (D soln )(V) = (1.0g/ml)(75.0ml) = 75.0g; m soln = m water + m KOH = = 85.0g q surr = C soln m total (T f T i ) 10.3kJ(1000J/1kJ) = (4.184J/g C)(85.0g)(T f 25.0) T f = 54.0 C HEAT of FORMATION, Δ f H : The heat, Δ f H, required to form 1mol of substance from its elements in their natural state under standard conditions (the means standard conditions: T = 25 C, P = 1bar, solns = 1M) Δ r H = Σn p Δ f H (products) - Σn r Δ f H (reactants) Δ f H = 0 for all elements in their natural state n p = # of product moles for each chemical, n r = # of reactant moles for each chemical Diatomic elements (memorize): F 2, Cl 2, Br 2, I 2, H 2, N 2, O 2 ; Liquid elements (memorize): Br 2, Hg; Gaseous elements (memorize): He, Ne, Ar, Kr, Rn, F 2, Cl 2, H 2, N 2, O 2 ; Rest are solids; Standard state for C (memorize): C graphite Example 1: Write Δ f H reaction for NH 4 ClO 4 (s). Answer: 1 / 2 N 2 (g) + 2H 2 (g) + 1 / 2 Cl 2 (g) + 2O 2 (g) NH 4 ClO 4 (s) must have phases shown; must form 1mol of product Example 2: Calculate Δ r H for 2CO(g) + O 2 (g) 2CO 2 (g) given Δ f H (CO) = kJ/mol and Δ f H (CO 2 ) = kJ/mol Answer: Δ r H = [2(Δ f H (CO2) )] [2(Δ f H (CO) ) + 1(Δ f H (O2) )] = [2(-393.5)] [2(-110.5) + 1(0)] = -566kJ (Note: Δ f H (O2(g)) = 0 element in natural state)

3 PHASE CHANGE + HEAT TRANSFER Example: 65.5g of iridium (Ir) is heated to 85.5 C and then placed on top of a block of ice at 0.00 C. After thermal equilibrium had been established, the temperature of the metal and ice were 0.00 C, and it was found that 2.24g of ice had melted. What is the specific heat capacity of the Ir? C s (H 2 O(s)) = 2.11J/g C, C s (H 2 O(l)) = 4.184J/g C, Δ fus H = 333J/g Answer: 0.133J/g C heat transfer + phase change; heat lost + heat gained = 0; for Ir: q = C s m(t f -T i ); for ice: q = Δ fus H(m); add the heats: C s (65.5)(0-85.5) + (333)(2.24) = 0; C s = 0; C s = ; C s = J/g C Fuel value: quantity of energy released when 1g of substance is burned to form CO 2 (g) and H 2 O(l) Caloric value: quantity of energy released in the body when 1g of substance is metabolized Energy density: quantity of energy released per unit volume of fuel Basal Metabolic Rate (BMR): minimum amount of energy to sustain a body 1. Ammonia burns in the presence of a copper catalyst to form nitrogen gas. 4NH 3 (g) + 3O 2 (g) 2N 2 (g) + 6H 2 O(g) Δ r H = 1270 kj/mol a. How much heat would be given off if 1.00mol of NH 3 was consumed? b. How much heat would be given off if 1.00g of NH 3 was consumed? 2. The overall reaction for the corrosion of iron by oxygen (rusting) is 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) When 10.0g of Fe is reacted, 73.9kJ are released. Calculate the Δ r H. 3. Hydrazine, N 2 H 4 (l), which is a liquid used as a rocket fuel, reacts with O 2 (g) to produce N 2 (g) and H 2 O(l). The reaction of 2.50grams of N 2 H 4 (l) with excess O 2 (g) gives off 48.5kJ of heat. Calculate the enthalpy change if three moles of hydrazine were combusted? a. 207kJ b. 621kJ c. 1870kJ d. 11.3kJ e. none of the above 4. Calculate the standard enthalpy change Δ r H for 2C(s) + H 2 (g) C 2 H 2 (g) given the individual reaction standard enthalpy changes Rxn I: C 2 H 2 (g) + 5 / 2 O 2 (g) 2CO 2 (g) + H 2 O(l) Δ r H = 1300 kj/mol Rxn II: C(s) + O 2 (g) CO 2 (g) Rxn III: H 2 (g) + 1 / 2 O 2 (g) H 2 O(l) 5. Calculate the standard enthalpy change Δ r H for Δ r H = 394 kj/mol Δ r H = 286 kj/mol N 2 H 4 (l) + CH 4 O(l) CH 2 O(g) + N 2 (g) + 3H 2 (g) given the individual reaction standard enthalpy changes Rxn I: N 2 H 4 (l) + H 2 (g) 2NH 3 (g) Δ r H = 9.0 kj/mol Rxn II: N 2 (g) + 3H 2 (g) 2NH 3 (g) Rxn III: CH 2 O(g) + H 2 (g) CH 4 O(l) 6. Calculate the standard enthalpy change Δ r H for Δ r H = 23.0 kj/mol Δ r H = 32.5 kj/mol 2H 2 O(l) 2H 2 (g) + O 2 (g) given the individual reaction standard enthalpy changes Rxn I: CH 3 COOH(l) + 2O 2 (g) 2CO 2 (g) + 2H 2 O(l) Δ r H = 1742 kj/mol

4 Rxn II: CO 2 (g) C(graphite)(s) + O 2 (g) Δ r H = kj/mol Rxn III: CH 3 COOH(l) 2C(graphite) + 2H 2 (g) + O 2 (g) Δ r H = kj/mol 7. When 4.00g of methane, CH 4, is burned in a constant volume bomb calorimeter (calorimeter constant = 2.25kJ/ C) with 250.ml H 2 O(l). After the reaction of the CH 4, the temperature of the calorimeter and contents increased from C to C. What is the ΔE for CH 4? D soln = 1.00g/ml; C soln = 4.184J/g C Hint: Start by writing the combustion reaction for CH 4. Assume the 4.00g of reactants or products are absorbed into the water. (Note: In bomb calorimeters you are calculating ΔE rather than Δ r H since the volume is constant but the calculation steps are the same as calculating Δ r H!) 8. When 5.00g CsOH was dissolved in 55.0ml of water in an insulated coffee cup, the temperature increased from 25.0 C to 34.5 C. What is the heat of solvation (Δ solvation H) for CsOH? The D soln = 1.00g/ml and C soln = 4.184J/g C. CsOH(s) CsOH(aq) g NaC 2 H 3 O. 2 3H 2 O (molar mass = 136g/mol) was dissolved in 45.5ml H 2 O(l) at 33.5 C according to the reaction: NaC 2 H 3 O. 2 3H 2 O(s) NaC 2 H 3 O 2 (aq) + 3H 2 O(l) Δ solvation H = 19.66kJ/mol What is the final temperature of the solution? The D soln = 1.00g/ml and C soln = 4.184J/g C ml of a 2.0M HCl solution at 0.0 C is mixed with 100.ml of a 2.0 M NaOH solution also at 0.0 C. If D soln = 1.00g/ml, C soln = 4.184J/g C, and Δ r H = -56kJ/mol, what is T f of the solution? 11. Write the Δ f H reaction that corresponds with each chemical listed below. a. CH 3 OH(l) b. NH 4 OH(s) c. Cu(NO 3 ) 2 (s) 12. The standard enthalpy change for the reaction below 1125kJ. Calculate the standard enthalpy of formation for SO 2 (g). 2H 2 S(g) + 3O 2 2H 2 O(l) + 2SO 2 (g) Δ f H for H 2 S(g) = -20.2kJ/mol, Δ f H for H 2 O(l) = kJ/mol, 13. Given the information below, what is the standard enthalpy of formation for KClO 3 (s)? 3 / 2 KClO 3 (s) 3 / 2 K(s) + 3 / 4 Cl 2 (g) + 9 / 4 O 2 (g) Δ r H = 597kJ/mol 14. The combustion of methanol takes place according to the reaction: 2CH 3 OH(l) + 3O 2 (g) 2CO 2 (g) + 4H 2 O(l) Compute Δ r H for the combustion of one mole of methanol under standard conditions and select the correct answer, given the following standard heats of formation: Δ f H for CH 3 OH(l) = kJ/mol Δ f H for CO 2 (g) = kJ/mol Δ f H for H 2 O(l) = kJ/mol Δ f H for CH 4 = kJ/mol Δ f H for CO(g) = kJ/mol Δ f H for H 2 O(g) = kJ/mol a kj/mol b kj/mol c kj/mol d kj/mol e kj/mol

5 15. a. Write an appropriate equation illustrating the standard enthalpy of formation for MnO 2 (s). b. Given the two reactions below, calculate Δ f H for MnO 2 (s). MnO 2 (s) MnO(s) + 1 / 2 O 2 (g) Δ r H = 136.0kJ/mol MnO 2 (s) + Mn(s) 2MnO(s) Δ r H = kJ/mol 16. When 25g of Ag(s) is combusted with O 2 (g) to form Ag 2 O(s), 3.597kJ are released. What is the standard enthalpy of formation for Ag 2 O? (Hint: Start by writing out Δ f H for Ag 2 O.) 17. When 2.750g of O 2 (g) reacts with Cr to form Cr 2 O 3 (s), 65.01kJ are released. What is the standard enthalpy of formation for Cr 2 O 3? (Hint: Start by writing out Δ f H for Cr 2 O 3.) (The next 4 questions are heat transfer+phase change.) g of vanadium (V) is heated to C and then placed in water at C. After thermal equilibrium had been established, the temperature of the V and water were C, and it was found that 1.22g of water had boiled. What is the specific heat capacity of the V? C s (H 2 O(l)) = 4.184J/g C, C s (H 2 O(g)) = 2.00J/g C, Δ vap H = 2256J/g g of titanium is cooled to C and placed on 12.75g ice at 0.00 C. After thermal equilibrium had been established, the temperature of the titanium and ice were 0.00 C, and 9.43g of ice remains. What is the specific heat capacity of titanium? C s (H 2 O(l)) = 4.184J/g C, C s (H 2 O(s)) = 2.11J/g C, Δ fus H = 333J/g (The next 2 questions are harder!) 20. If 50.0g metal (C = 4.00J/g C) initially at 80.0 C is placed in 25.0g water at 55.0 C, what is the final temperature and how many grams of water, if any, freezes? The heat capacity of water = 4.184J/g C, heat capacity of ice = 2.11J/g C, and heat of fusion for water is 333J/g. 21. A beaker contains 50g of water (C = 4.184J/g C) at 20 C and a 10gram piece of metal (C = 5.75J/g C) at 450 C is added to it. How many grams of water boiled and what is the final temperature in the beaker? The heat of fusion for water is 333J/g and the heat of vaporization is 2256J/g. ANSWERS 1. a. -318kJ {(4mol NH 3 )/-1270kJ = (1mol NH 3 )/x; x = -318kJ} b kj {(4mol NH 3 )/-1270kJ = (1g NH 3 )/x; (68.14g NH 3 )/-1270kJ = (1g NH 3 )/x; x = -18.6kJ} kj {(4mol Fe)/x = (10g Fe)/(-73.9); (223.4g Fe)/x = (10g Fe)/(-73.9); x = kJ} 3. c {(3mol N 2 H 4 )/x = (2.5g N 2 H 4 )/(-48.5); (96.16g N 2 H 4 )/x = (2.5g N 2 H 4 )/(-48.5); x = -1865kJ} kj/mol {multiply Rxn II by 2; no change to Rxn III; reverse Rxn I; than add} kJ/mol {no change to Rxn I; reverse Rxn III; reverse Rxn II} kJ/mol {Rxn I: reverse; Rxn III: no change; by inspection of Rxn I and Rxn III Rxn II: reverse and multiply by 2} kJ {write reaction: CH 4 + 2O 2 CO 2 + 2H 2 O; Step1: q surr = q calorimeter + q H2O ; q surr = C cal ΔT + C soln m mix ΔT; q surr = (2.25kJ/ C)(1000J/1kJ)( ) + (4.184)[(250ml)(1.00g/ml) g]( ) = J J = 200,255J; Step 2: q surr = -q sys ; q sys = -200,255J; Step 3: Proportionality: 4.00g CH 4 /-200,255J = 1mol CH 4 /ΔE; convert 1 mol CH 4 to grams: 1mol CH 4 x (16.04g CH 4 /1mol CH 4 ) = 16.04g CH 4 ; 4.00g CH 4 /-200,255J = 16.04gCH 4 /ΔE; 4.00ΔE = -3,212,090J; ΔE = J x 1kJ/1000J = -803kJ} kJ/mol {Step 1: q surr = C soln m soln (T f - T i ); q surr = (4.184J/gC)(5.00g g)( ) = 2385J; Step 2: q surr = -q sys ; q sys = -2385J; Step 3: Proportionality: 1mol CsOH/xJ = 5.00g CsOH/-2385J; 149.9g CsOH/xJ = 5.00g CsOH/-2385J; 5x = ; x = J = -71.5kJ = ΔH}

6 C {Step 1: Proportionality: 1mol NaC 2 H 3 O 2. 3H2 O(s)/19.66J = 14.20g NaC 2 H 3 O 2. 3H2 O(s)/x; 136g NaC 2 H 3 O 2. 3H2 O(s)/19.66J = 14.20g NaC 2 H 3 O 2. 3H2 O(s)/x; x = 2.053kJ; Step 2: q sys = -q surr ; q surr = kJ; Step 3: q surr = C soln m soln (T f - T i ); kJ(1000J/1kJ) = (4.184J/gC)[(45.5ml)(1.00g/ml) g)(T f 33.5); -2053J = T f ; = T f ; T f = C} 10. a. 8.9 C {First, write reaction: HCl(aq) + NaOH(aq) H 2 O(l) + NaCl(aq) with a Δ r H =-56kJ/mol; find moles that react: limiting reagent problem: M HCl x vol HCl = mol HCl = (2.0M)(0.05L) = 0.1mol HCl; M NaOH x vol NaOH = mol NaOH = (2.0M)(0.1L) = 0.2mol NaOH; ratio is 1:1 HCl is the limiting reagent; Step 1: Proportionality: how much heat evolved when 0.1mol reacts: 1mol HCl/-56kJ = 0.1mol HCl/x x = -5.6kJ = -5600J; q sys = -5600J; Step 2: q surr = -q sys ; q surr = 5600J; Step 3: q surr = CmΔT; the heat goes into the 2 solutions which total 50ml + 100ml = 150ml; convert to mass with density: D = mass/vol mass = D x vol = 1.00g/ml x 150. ml = 150. g solution; use q surr = C soln m soln (T f - T i ); 5600J = (4.184)(150.)(T f 0.00); 5600 = 627.6T f ; T f = 8.92 C} 11. a. C(graphite) + 2H 2 (g) + 1 / 2 O 2 (g) CH 3 OH(l) b. 1 / 2 N 2 (g) + 1 / 2 O 2 (g) + 5 / 2 H 2 (g) NH 4 OH(s) c. Cu(s) + N 2 (g) + 3O 2 (g) Cu(NO 3 ) 2 (s) kJ/mol {-1125 = [2(-285.8) + 2(Δ f H SO2 )] [2(-20.2) +3(0)]; solve for Δ f H (SO2) } kJ {write Δ f H for KClO 3 : K(s) + 1 / 2 Cl 2 (g) + 3 / 2 O 2 (g) KClO 3 (s); take rxn given in problem and reverse and multiply by 2/3; do the same with Δ r H and that is the value for Δ f H } 14. b {Δ r H = [2(-393.5) + 4(-285.6)] [2(-238.5) + 3(0)] = kJ this is for 2 mol CH 3 OH; (2mol CH 3 OH)/( ) = (1mol CH 3 OH)/x; x = kJ} 15. a. Mn(s) + O 2 (g) MnO 2 (s) b kj/mol kJ/mol {2Ag(s) + 1 / 2 O 2 (g) Ag 2 O(s); (25g Ag)/-3.597kJ = (2mol Ag)/x; x = kJ} kJ {write Δ f H for Cr 2 O 3 : 2Cr(s) + 3 / 2 O 2 (g) Cr 2 O 3 (s); do a proportionality: 2.75g O 2 /-65.01kJ = 1.5mol O 2 /x; mol O 2 /-65.01kJ = 1.5mol O 2 /x; solve for x: x = kJ} J/g C {heat transfer + phase change; heat lost + heat gained = 0; for V: q = C s m(t f -T i ); for water: q = Δ vap H(m); C s (45.0)( ) + (2256)(1.22) = 0; -5625C s = 0; 5625C s = 2752; C s = J/g C} J/g C {heat transfer + phase change; heat lost + heat gained = 0; for Ti: q = C s m(t f -T i ); for water: q = Δ fus H(m); 12.75g ice 9.43g ice = 3.32g ice melted; C s (50.0g)( C) + (333J/g)(3.32g) = 0; C s = 0; C s = ; C s = J/g C} C and all the water freezes {q heat metal + q cool water + q freeze water + q cool ice = 0; started by assuming T f = 0 C: q cool water = (4.184)(25.00)( ) = -5753J; q heat metal to 0 C = (4.00)(50.0)(0 (-80.0)) = 16000J; since the metal will cool the water down to 0 C and the metal at this time will still be lower than 0 C, some water freezes; assume all water freezes: q freeze water = (-333)(25.0) = -8325J; since cooling and freezing all the water yields (-8325) = J, and it requires 16000J to heat the metal to 0 C, the ice then cools below 0 C q heat metal + q cool water + q freeze water + q cool ice = (4.00)(50.0)(T f (-80.0)) + (-5753) + (-8325) + (2.11)(25.0)(T f 0) = 0; 200T f T f = 0; T f = -1922; T f = C} g boils; T f = 100 C {q cool metal + q heat water + q boil water + q heat H2O(g) = 0; q heat water to 100 C = (4.184)(50)(100-20) = 16736J needed; assume metal cools to 100 C: q metal cooling to 100 C = (5.75)(10)( ) = J; sufficient to raise water to 100 C and some/all water boils; assume all the water boils: q boil water = 2256(50) = J; not sufficient amount of heat from the metal to boil all the water so only some water boils and T f = 100 C q cool metal + q heat water + q boil water = 0; (2256)(x); x = 4.05g boils; T f = 100 C}