Calorimetry+Problems

m,water= .5kg Ti=20 C Tf=80 C m, copper= 1kg Cwater= 4.18j/kgC Ccopper= 0.385kj/KgC
 * 1**. How much heat is needed to raise the temperature of the copper kettle and its contenets to 80 C?

Delta T = Tf-Ti = 80 - 20 = 60 C

Q = mcT

Qwater= (0.5)(4.18)(60) Qwater= 125.4kj

Qcopper = (1)(0.385)(60) Qcopper = 23.1kj

Qtotal= Qwater + Qcopper Qtotal= 125.4 + 23.1 Qtotal= 148.5kj

The total heat required is **148.5kj**.

2. The heat of neutralization of HCl with NaOH is 57 kJ/mol (for every mole of NaOH which is neutralized by the HCl, 57 kJ of heat is released).

a**) How much heat is released by mixing 50 mL of 1.0 M acid with 50 mL of 1.0 M base?**

50 mL NaOH/x mol (times) 1000ml NaOH/1 mol (cross-multiply) 50 = 1000x x = 0.05 x = 0.05 moles

1 mol NaOH/57 kJ (times) 0.05 mol NaOH/ x kJ (cross-multiply) x = (0.05)(57) x = 2.85 x = 2.85 kJ

Therefore 2.8 kJ are released.

q = mc ∆T q = 2.85kJ (x 1000) = 2850 J m = 50 mL x (1g / mL) = 50 g (x2) = 100 g c = 4.18J/g C
 * b) If both solutions were initially at 20 degrees Celsius, what will be the final temperature of the mixture?**

2850 = (100) (4.18__)__ ∆T 2850 = 418∆T 2850/418 = 418∆T/418 ∆T = 6.82 ∆T = 6.8 degrees Celsius 20 degrees Celsius (initial temp.) + 6.8 degrees Celsius (change in temp) = 26. 8 degrees Celsius

Therefore temperature of the mixture (or temp. of surroundings since you're referring to the water) is 26.8 degrees Celsius.

3. **The burning of 1 mol of methane releases 900kJ. How much methane must be burned to heat 100L of water for a bath from a starting temperature of 20 C to a final temperature of 45 C?** c(H20)=4.18 J/g*C

q=? q(CH4)=900 kJ/mol n(CH4)=? m(CH4)=?

V(H2O)=100 mL m(H2O)=100 g

T(f)=45 C T(i)=20 C ∆T=25 C

q=mc ∆T q=(100)(4.18)(25) q=10 450 kJ

__1 mol__ =__ ........ n ......... __ . 900kJ ........ 10 450kJ n=11.61111...mol

m(CH4)=mm(CH4)*n ............ =(16)(11.6111...) ............ =185.78 g Therefore, 185.78g of methane must be burnt to heat the bath water 25 C!

DON'T BOTHER FORMATTING WITH SPACES. IT JUST GETS MESSED UP. DX

4. Givens per gram of wax= 52.5 J of energy released (Q) Density of water= 1g/ml c of water= 4.18 J/Cg D=m/v 1m3= 1kJ Vwater =0.5x10^3 L= 0.5 g= m of water

Required how many grams of wax required? and how much Q is needed?

Qsurr= 0.5g(4.18J/Cg)(20C)= 41.8 J

52.5/1g = 41.8J/x x=( 41.8J(1g))/ 52.5 kJ x= - 0.796g/ J x= 796g Solution: The mass of wax required to raise the temp. of the water by 20 C is 796g.

5. The standard enthalpy of solution of a substance is the enthalpy change occuring when one mole of that substance is dissolved in a large volume of water. How much of a temperature ecrease will be produced by dissolving 1.0g of this ammunium nitrate in 50mL of water? The heat of solution of NH4NO3 is 25.8 kJ/mol. G. Molar mass of NH4NO3 = 80g/mol 50mL water = 50g water Head of solution of NH4NO3 = 25.8kJ/mol = 25800J/m c = 4.18J/gC

R. Change in temperature

A. Q = mcT Q = Heat of solution x # of mols of NH4NO3 Moles = Mass/Molar mass

S. #of moles = 1g / 80g/mol = 0.0125 mols

Q = 0.0125 mols x 25800J/mol =322.5J

322.5 = (50)(4.18)T T = 1.5degrees

P. The water will increase by 1.5 degrees (the NH4NO3 will drop by 1.5 degrees).

6. **The standard enthalpy of solution of NaOH is -42.7kJ/mol and the the standard enthalpy of neutralization of NaOH is -57.3kJ/mol. Calculate the temperature rise produced by dissolving 1g of NaOH in:** __a) 100mL of water__ Given: q=-42.7kJ/mol, m=100g, c=4.18J/g °C Required:ΔT the enthalpy of NaOH has to be converted to kJ per gram instead of moles: __-42.7kJ/mol__=-1.07kJ/g 40g/mol

Equation: q sys=mcΔ Tsys

-1.0675kJ/g=100g(4.18J/g°C)Δ Tsys

__-1067.5J/g__ = Δ Tsys 100g (4.18J/g°C)

-2.56°C=ΔTsys

ΔTsys= - ΔTsurr

-2.56°C= -ΔTsurr

2.56°C= ΔTsurr Therefore, 100mL of water gains 2.6°C when 1g of NaOH is dissolved in it.

__b) 100mL of 0.15M HCl__ Part 1: Since all of the NaOH dissolves, use the amount of energy released from part a) (-1.0675 kJ)

Part 2: You have to add to this the change in temperature caused by the NaOH neutralizing. use stoichiometry to determine how much NaOH actually neutralizes

NaOH + HCl = NaCl + H20 most of you probably know this but 0.15M HCl means that there is a concentration of HCl of 0.15 mol/L in the solution, therefore, there is also a concentration of 0.15mol/L of NaOH to be completely neutralized Find the Number of moles of NaOH 0.15mol/L x 0.1L = 0.0015 mol

Determine the amount of energy released by neutralization (0.0015mol)(-57.3 kJ/mol) = -0.8595kJ

the Total amount of energy given off is the two energies added together Qsys = (-1.0675 kJ) + (-0.8595kJ) Qsys = -1.9265 kJ

-1926.5 J = (100g) (4.18J/g°C) <span style="font-family: 'Calibri','sans-serif'; font-size: 11pt; line-height: 115%;">ΔT __-1926.5 J__ = <span style="font-family: 'Calibri','sans-serif'; font-size: 11pt; line-height: 115%;">ΔT 100g(4.18J/g<span style="font-family: 'Calibri','sans-serif'; font-size: 11pt; line-height: 115%;">°C) Therefore the change in temperature when 1g of NaOH is neutralized with 0.15M HCl is **-5<span style="font-family: 'Calibri','sans-serif'; font-size: 11pt; line-height: 115%;">°C. ** (I know this is not the right answer as far as the sheet goes but Leslie confirmed this is the right answer)
 * -4.608<span style="font-family: 'Calibri','sans-serif'; font-size: 11pt; line-height: 115%;">°C = <span style="font-family: 'Calibri','sans-serif'; font-size: 11pt; line-height: 115%;">ΔT **

c) 100mL of 0.15M H2SO4

Same idea as b) Stoichiometry H2SO4 + 2 NaOH = Na2SO4 + 2 H2O Therefore there is 2x as much NaOH required as H2SO4 0.15MNaOH x 2 = 0.3M NaOH __1.0g NaOH__ = number of moles 40g/mol 0.025mol = number of moles 0.3mol/L x 0.1 L = 0.03 mol of NaOH neutralized (theoretically) therefore all of the NaOH is Neutralized 0.025mol x -57.3kJ/mol = -1.4325kJ

Use energy released from a)

Qtotal = -1.067kJ + (-1.4325kJ) Qtotal = -2.4995kJ

- 2499.5 J = (100g) (4.18 J/g<span style="font-family: 'Calibri','sans-serif'; font-size: 11pt; line-height: 115%;">°C) <span style="font-family: 'Calibri','sans-serif'; font-size: 11pt; line-height: 115%;">ΔT __- 2499.5 J__ = <span style="font-family: 'Calibri','sans-serif'; font-size: 11pt; line-height: 115%;">ΔT 100g(4.18 J/g<span style="font-family: 'Calibri','sans-serif'; font-size: 11pt; line-height: 115%;">°C)
 * -5.98<span style="font-family: 'Calibri','sans-serif'; font-size: 11pt; line-height: 115%;">°C = <span style="font-family: 'Calibri','sans-serif'; font-size: 11pt; line-height: 115%;">ΔT **

Therefore the change in temperature when 1g of NaOH is neutralized with 0.15M H2SO4 in 100mL of water is **-6<span style="font-family: 'Calibri','sans-serif'; font-size: 11pt; line-height: 115%;">°C. ** = =

=
7. A man is immersed in a tub containing 60.0L of Water. The heat of his body raises the temperature of the water from 30.0°C to 31.5°C in one hour: a) At what rate is the man giving off heat? b) How many kJ would he give off in a day?====== <span style="font-family: Calibri,sans-serif; font-size: 15px; line-height: 17px;">**__a)__** __Given:__ <span style="font-family: arial,helvetica,sans-serif; font-size: 13px; font-weight: normal; line-height: 20px;"><span style="font-family: Calibri,sans-serif; font-size: 11pt; line-height: 17px;">//Δ// //T// = 1.5<span style="font-family: Calibri,sans-serif; font-size: 15px; line-height: 17px;">°C Mass of H2O = 60 L = 60 kg = 60000 g __Needed:__ //Q of H2O per Hour// __Analysis:__ //Q = (mass)*(change in temperature)*(specific heat capacity)//

//Q(H2O)// = (60000 g)(1.5°C)(4.18 J/g°C)(1/h) = 376200 J/h = 376.2 kJ/h

//Since the change in T (1.5°C) occurred in a period of an hour, 376.2 kJ is therefore the amount of heat emitted in an hour by this man.//


 * __b)__**

__Given:__ //Q(H2O)/Hour// = 376.2 kJ /h //t// = 1 day = 24 hours __Needed:__ //Q(H2O)/day// __Analysis:__ //Q/h * t = Q/day//

376.2 kJ/h * 24 hours/1 day = 9028.8 kJ/day

//Since the amount of heat emitted in an hour is 376.2 kJ, 9028.8 kJ is therefore the amount of heat emitted by this man in a day.//