CHEM1101 2014-J-8 June 2014

• A 1.0 kg sample of copper metal is heated to 100.0 °C. The copper sample is immersed in a volume of water initially at 25.0 °C. What volume of water is required so that the final temperature of the copper is 40.0 °C? Show all working.

Data: Specific heat capacity of Cu(s) is 0.39 J K–1 g–1. Specific heat capacity of H2O(l) is 4.184 J K–1 g–1. The density of water is 1.0 g mL–1.

Marks 3

The copper cools from 100.0 oC to 40.0 oC so ΔT = -60.0 K. As the specific heat capacity of Cu(s) is 0.39 J K-1 g-1, the heat lost by 1.0 kg during this is given by:

q = mC ΔT = (1.0 × 103 g) × (0.39 J K-1 g-1) × (-60.0 K) = -23000 J

This heat is gained by the water which is warmed from 25.0 oC to 40.0 oC – a change of 15.0 K. As the specific heat capacity of water is 4.184 J K-1 g-1 and a temperature change of 15.0 K is required:

q = mC ΔT = m × (4.184 J K-1 g-1) × (15.0 K) = 23000 J

m = 370 g

The density of water is 1.0 g mL-1 so the volume required is: volume = mass / density = 370 g / 1.0 g mL-1 = 370 mL

Answer: 370 mL

THE REMAINDER OF THIS PAGE IS FOR ROUGH WORKING ONLY.

CHEM1101 2014-N-9 November 2014

• Lead shot was traditionally made by dropping molten lead into a tank of water. A piece of lead, initially at 327 °C is dropped into 200.0 mL of water raising its temperature from 25 to 35 °C. What was the weight of the lead? Data: Specific heat capacity of Pb is 0.126 J K–1g–1

Specific heat capacity of H2O(l) is 4.184 J K–1g–1 The density of water is 1.0 g mL–1

Marks 2

200.0 mL of water corresponds to 200.0 g. When this quantity warms from 25 oC to 35 oC, the heat change is:

qwater = mCΔT = (200.0 g) × (4.184 J g-1 K-1) × ((35 - 25) K) = 8400 J

The mass of lead is x g, the heat change when it cools from 35 oC to 327 oC is:

qlead = mCΔT = (x) × (0.126 J g-1 K-1) × ((35 - 327) K) = -36.7x J

The heat gained by the water is lost by the lead so these two quantities must be equal in magnitude:

qwater = - qlead

8400 J = 36.7x J

x = 230 g

Answer: 230 g

THE REMAINDER OF THIS PAGE IS FOR ROUGH WORKING ONLY.

CHEM1101 2013-J-8 June 2013

• 1.00 L of water is heated to 95 oC and then solid copper, initially at 25 oC, is immersed in it. What mass of copper was added if the final temperature of the water was 84 oC? Show all working.

Data: Specific heat capacity of Cu(s) is 0.39 J g–1 K–1. Specific heat capacity of H2O(l) is 4.184 J g–1 K–1. The density of water is 1.0 g mL–1.

Marks 3

1.00 L of water corresponds to 1.00 × 103 g. When this quantity cools from 95 oC to 84 oC, the heat change is:

qwater = mCΔT = (1.00 × 103 g) × (4.184 J g-1 K-1) × ((84 - 95) K) = -46 × 103 J

If the mass of copper used is x g, the heat change when it warms from 25 oC to 84 oC is:

qcopper = mCΔT = (x) × (0.39 J g-1 K-1) × ((84 - 25) K) = +23x J

The heat lost by the water is gained by the copper so these two quantities must be equal in magnitude:

23x = 46 × 103 so x = 2.0 kg

Answer: 2.0 kg

CHEM1101 2012-J-9 June 2012

• A 2.5 kg block of aluminium is heated to 80.0 °C and then placed into a thermally insulated water bath consisting of 10.0 L of water at 25.0 °C. Calculate the final temperature of the water once equilibrium has been reached. Show all working.

Data: Specific heat capacity of Al(s) is 0.900 J g–1 K–1. Specific heat capacity of H2O(l) is 4.184 J g–1 K–1. The density of water is 1 g mL–1.

Marks 3

The aluminium will cool down and the water will heat up when the two are mixed. The final temperature, Tf, will be the same for both. For the aluminium, qaluminium = m c ΔT = (2500 g) × (0.900 J g-1 K-1) × ΔTaluminium

For the water, qwater = m c ΔT = (10.0 × 103 g) × (4.18 J g-1 K-1) × ΔTwater

As the heat lost by the aluminium is gained by the water, qwater = -qaluminium: (10.0 × 103 g) × (4.18 J g-1 K-1) × ΔTwater = - (2500 g) × (0.900 J g-1 K-1) × ΔTaluminium or 41800 × ΔTwater = - 2250 × ΔTaluminium

Using ΔTwater = (Tf – 25.0) oC and ΔTaluminium = (Tf – 80.0) oC gives:

Tf = 27.8 oC

Answer: 27.8 oC

CHEM1101 2012-N-10 November 2012

• How many 2.0 L casks of wine and/or juices can be cooled on a hot Sydney day from 30 oC to a drinkable 10 oC with one 10.0 kg bag of ice taken from a freezer at –30 oC? Assume the specific heat of the wine and/or juice is the same as that of water, that the cardboard or plastic containers have negligible heat capacity, and that the density of the wine and juices is 1.0 g mL–1.

ΔfusH(H2O) = 6.0 kJ mol–1; Cp°(H2O(s)) = 2.2 J K–1 g–1; Cp°(H2O(l)) = 4.2 J K–1 g–1

Marks 6

The bag of ice will undergo three process in being raised from -30 oC to +10 oC:

(a) The ice will be heated from -30 oC to 0 oC for which ΔT = +30 K:

q(a) = mCp°ΔT = (10.0 × 103 g) × (2.2 J K-1 g-1) × (30 K) = 6.60 × 102 kJ

(b) The ice will melt. As the molar mass of H2O is (16.00 + 2 × 1.008) g mol-1 or 18.016 g mol-1. The number of moles of ice that will be melted is: number of moles = mass / molar mass = (10.0 × 103 g) / (18.016 g mol-1) = 555 mol As ΔfusH(H2O) = 6.0 kJ mol–1, the amount of energy required to melt this amount is: q(b) = (555 mol) × (6.0 kJ mol-1) = 3.33 × 103 kJ

(c) The melted ice will be heated from 0 oC to +10 oC for which ΔT = +10 K:

q(c) = mCp°ΔT = (10.0 × 103 g) × (4.2 J K-1 g-1) × (10 K) = 4.20 × 102 kJ

Overall, the heat absorbed by the ice is therefore: qice = q(a) + q(b) + q(c) = (6.60 × 102 kJ) + (3.33 × 103 kJ) + (4.20 × 102 kJ) = 4.4 × 103 kJ This heat is taken from the wine casks so: qwine = -4.41 × 103 kJ The mass of each cask is 2.0 kg and the required temperature change is from +30 oC to +10 oC for which ΔT = -20 K. If x casks are cooled: qwine = mCp°ΔT = (2.0 × 103 g) × x × (4.2 J K-1 g-1) × (-20 K) = -4.41 × 103 kJ Hence: x = 26.3 i.e. 26 casks can be cooled.

Answer: 26 casks

ANSWER CONTINUES ON THE NEXT PAGE

CHEM1101 2012-N-10 November 2012

What other assumption have you made in your calculation?

It has been assumed that all the heat that melts the ice and then warms the water comes solely from the casks being cooled. This clearly will not be the case - the system is exposed to the atmosphere so that people can access the drinks.

CHEM1101 2010-J-15 June 2010

• Consider two blocks of steel: block A is 1.00 kg and block B is 600. g. Both blocks start from the same temperature and are heated so that 600. J flows into each of the blocks in the form of heat. What is the final difference in temperature, TA-TB, between block A and block B. The specific heat of steel is 0.460 J g–1 K–1. Show all working.

2

The specific heat capacity C can be used to calculate the increase in temperature, ∆T from a heat change, q as:

q = m C ∆T

where m is the mass of the substance.

For block A, m = 1.00 × 103 g so that the temperature increase is:

∆T = q / mC = (600. J) / (1.00 × 103 g) × (0.460 J g-1 K -1) = 1.30 K

For block A, m = 600. 103 g so that the temperature increase is:

∆T = q / mC = (600. J) / (600. g) × (0.460 J g-1 K -1) = 2.17 K

The final temperature difference, TA-TB, is therefore (1.30 – 2.17) K = -0.87 K

Note that block B is lighter so ends up as the hotter of the two.

Answer: –0.87 K

CHEM1101 2010-N-7 November 2010

• Water solutions of NaOH (100.0 mL, 2.0 M) and HCl (100.0 mL, 2.0 M), both at 24.6 °C, were mixed together in a coffee cup calorimeter. The temperature of the solution rose to 38.0 °C during the reaction process. Write a balanced ionic equation to describe the reaction in the calorimeter.

Marks5

H+(aq) + OH–(aq) → H2O(l)

Is the process an endothermic or exothermic reaction? exothermic

Assuming a perfect calorimeter, determine the standard enthalpy change for the neutralisation reaction. Assume the density of water is 1.00 g mL–1. The heat capacity of water is 4.18 J g–1 K–1.

The temperature rise is (38.0 – 24.6) °C = 13.4 °C. The temperature difference corresponds to 13.4 K.

After mixing, the total volume of solution is (100.0 + 100.0) mL = 200.0 mL. With a density of 1.00 g mL-1, this corresponds to 200. g.

The heat change is given by q = m C ΔT where C is the specific heat capacity. Hence:

q = (200. g) × (4.18 J g-1 K-1) × (13.4 K) = 11200 J

100.0 mL of 2.0 M NaOH (or HCl) contains:

number of moles = concentration × volume

= (2.0 mol L-1) × (0.1000 L) = 0.20 mol

As 11200 J are released by 0.20 mol, the is the molar enthalpy change is therefore:

ΔH = -(11200 J) / (0.20 mol) = 56000 J mol-1 = 56 kJ mol-1

Answer: -56 kJ mol-1

CHEM1101 2009-J-12 June 2009

A radiator generates 150 J to heat up air inside a sealed container with volume of

2.00 L and initially at 25 C and atmospheric pressure. What will be the pressure

inside the container after heating?

Assume that air is composed of 80 % nitrogen and 20 % oxygen by volume.

Molar heat capacities: N2 29.14 J K–1

mol–1

and O2 29.38 J K–1

mol–1

Marks

5

Using the ideal gas law, the number of moles in 2.00 L of air at 25 °C ( = 298.0 K)

at atmospheric pressure is:

n = 𝑷𝑽

𝑹𝑻 =

𝟏.𝟎𝟎 𝐚𝐭𝐦 (𝟐.𝟎𝟎 𝐋)

𝟎.𝟎𝟖𝟐𝟎𝟔 𝐋 𝐚𝐭𝐦 𝐊−𝟏 𝐦𝐨𝐥−𝟏 (𝟐𝟗𝟖.𝟎 𝐊) = 0.0818 mol

The temperature change, ΔT, is related to the heat change, q, through,

q = n Cp ΔT or ΔT = 𝒒

𝒏𝑪𝒑

The ideal gas law shows that the number of moles is directly proportional to the

volume. The molar heat capacity of air is therefore the average of individual

molar heat capacities:

Cp(air) = 0.80Cp(N2) + 0.20Cp(O2) = [(0.80×29.14) + (0.20×29.38)] J K-1

mol-1

= 29.19 J K-1

mol-1

Hence,

ΔT = 𝒒

𝒏𝑪𝒑 =

𝟏𝟓𝟎 𝐉

𝟎.𝟎𝟖𝟏𝟖 𝐦𝐨𝐥 (𝟐𝟗.𝟏𝟗 𝐉 𝐊−𝟏 𝐦𝐨𝐥−𝟏) = 62.8 K

As the initial temperature is 298 K and the air is being heated, the final

temperature is (298.0 + 62.8) K = 360.8 K.

Finally, at this temperature, the ideal gas gives the pressure as:

P = 𝒏𝑹𝑻

𝑽 =

𝟎.𝟎𝟖𝟏𝟖 𝐦𝐨𝐥 𝟎.𝟎𝟖𝟐𝟎𝟔 𝐋 𝐚𝐭𝐦 𝐊−𝟏 𝐦𝐨𝐥−𝟏 (𝟑𝟔𝟎.𝟖 𝐊)

(𝟐.𝟎𝟎 𝐋) = 1.21 atm

Pressure: 1.21 atm

If this heated air is injected into a balloon, it will rise. Use the ideal gas equation to

explain why this happens.

From the ideal gas law, 𝒏

𝑽 =

𝑷

𝑹𝑻 . As the density is proportional to

𝒏

𝑽 , this shows

that the density is inversely proportional to the temperature. The heated air has a

lower density than the air in the atmosphere.

CHEM1101 2006-J-6 June 2006

• At room temperature and pressure (RTP), 1 mole of an ideal gas occupies 24.45 L. Calculate the molar volume of the same ideal gas in the stratosphere, where the pressure is 0.020 atm and the temperature is 200 K.

Marks 2

Using the ideal gas equation, PV = nRT, the volume of 1 mole of gas at P = 0.020 atm and T = 200 K is:

V = 2nRT 1×(0.08206)×(200)= =820L = 8.2×10 L

P (0.020)

Answer: 820 L= 8.2 × 102 L

• Two blocks of metal, as shown in the table below, are placed in intimate contact in an insulated environment.

4

Metal Iron Copper

Mass (g) 30.0 20.0

Initial T (°C) 0.0 100.0

c (J g−1 K−1) 0.450 0.387

In which direction will the heat flow? Write “from Fe to Cu” or “from Cu to Fe”.

As the copper is at a higher initial temperature, heat will flow from Cu to Fe

What is the final temperature of the system?

Heat will flow until the temperatures are equal. The heat lost by the copper is equal to the heat gained by the iron. The heat is related to the temperature change by the equation, q = m × c × ∆T. The heat lost by the copper in going from 100.0 °C to the final temperature Tf is:

q = mCu × cCu × ∆T = 0.387 × 20.0 × (100.0 – Tf).

The heat gained by the iron in going from 0.0 °C to the final temperature Tf is:

q = mFe × cFe × ∆T = 0.450 × 30.0 × (Tf – 0.0).

Setting the two heat changes to be equal gives:

0.387 × 20.0 × (100.0 – Tf) = 0.450 × 30.0 × (Tf – 0.0) or 774 – 7.74Tf = 13.5 Tf Hence, Tf = 36.4 °C

Answer: 36.4 °C

CHEM1101 2005-J-7 June 2005

Methane, CH4, represents an increasingly important fuel. Write the balanced chemical

equation for the combustion of methane.

Marks

4

CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

Calculate the mass of CO2 that would be produced by the combustion of 1.00 kg of

methane.

The molar mass of CH4 is (12.01 (C) + 4 1.008 (H)) g mol-1

= 16.042 g mol-1

.

The number of moles of CH4 in 1.00 kg is therefore:

number of moles = 3

1

1.00 10 g

16.042gmol

= 62.3 mol

In the reaction, 1 mol of CH4 leads to 1 mol of CO2. Hence 62.3 mol of CO2 will

be emitted from this number of moles of CH4. The molar mass of CO2 is (12.01

(C) + 2 16.00 (O)) g mol-1

= 44.01 g mol-1

. The mass of 62.3 mol is therefore:

mass of CO2 = (62.3 mol) (44.01 g mol-1

) = 2740 g = 2.74 kg

Answer: 2.74 kg

Calculate the volume of CO2 produced at 0 C and 1 atm.

Assuming that CO2 behaves like an ideal gas, the volume is given by PV = nRT.

Using n = 62.3 mol, T = 0 C = 273 K and P = 1 atm:

V = 1 1

(62.3mol) (0.08206LatmK mol ) (273K)

(1atm)

nRT

P

= 1400 L = 1.40 103 L

(As the pressure has been used in atmospheres, R = 0.08206 L atm K-1

mol-1

has

been used leading to the volume in litres).

Answer: 1.40 10

3 L

In an inefficient combustion reaction some of the methane gas may escape into the

atmosphere, thereby decreasing the amount of CO2 produced. Would such a leakage

lead to a greater or lesser enhancement of the Greenhouse Effect? Why?

Methane is a much more efficient greenhouse gas than carbon dioxide so the

Greenhouse Effect will be enhanced by release of CH4 instead of CO2.

CHEM1101 2004-N-7 November 2004

Two students determined the specific heat capacity of steel as follows. Student 1

poured 200 mL of recently boiled water into a styrofoam cup calorimeter and

measured the temperature of the water to be 90.0 °C. A steel teaspoon (25 g) at room

temperature (20.0 °C) was placed into the cup, where it was completely submerged.

After equilibrium was presumably reached, the temperature was measured to be

88.7 °C. Student 2 took the same cup and filled it with 200 mL of water at room

temperature (20.0 °C) and placed the same teaspoon from the freezer (– 40.0 °C) into

the cup. After equilibrium was presumably reached, the temperature recorded was

19.2 °C. Determine the specific heat capacity of steel using the experimental data of

each student. The specific heat capacity of water is 4.184 J g–1

K–1

. Assume the

density of water is 1.00 g mL–1

at all temperatures.

Marks

5

The specific heat capacity is related to the heat and temperature change by:

q = c m ΔT

In student 1’s experiment, the heat lost by the water is gained by the spoon. The

temperature of the water changes by (90.0 – 88.7) °C = 1.3 °C. The temperature

of the spoon changed by (88.7 – 20.0) °C = 68.7 °C. Equating the heat changes

gives:

cwater mwater ΔTwater = csteel msteel ΔTsteel

(4.184 J g-1

K-1

) (200 g) (1.3 K) = csteel (25.0 g) (68.7 K)

csteel = 0.63 J g-1

K-1

In student 2’s experiment, the heat lost by the water is gained by the spoon. The

temperature of the water changes by (20.0 – 19.2) °C = 0.8 °C. The temperature

of the spoon changed by (-40.0 – 19.2) °C = 59.2 °C. Equating the heat changes

gives:

cwater mwater ΔTwater = csteel msteel ΔTsteel

(4.184 J g-1

mol-1

) (200 g) (0.8 K) = csteel (25.0 g) (59.2 °C)

csteel = 0.45 J g-1

K-1

Student 1: 0.63 J g-1

K-1

Student 2: 0.45 J g-1

K-1

Which student's value is more accurate? Give reasons for your answer.

The largest source of error is likely to be lose of heat (student 1) or gain of heat

(student 2) from the surroundings. As student 2’s experiment is carried out with

water close to room temperature, this effect is likely to be much smaller than for

student 1 where the water is at 90 °C.