Physics JC1 2009
Tutorial 6: Thermal Physics
Use of Specific Heat Capacity and Specific Latent Heat
1. N94/2/7
The specific latent heat of vaporization of water at 28 °C is 2.3 x 106 J kg-1. It has been estimated that 1.2 x 1012 m3 of water is evaporated per day from the Earth’s surface. Given that the density of water is 1.0 x 103 kg m-3, calculate the energy required to evaporate this volume of water. [2.8 x 1021 J]
Solution
Since the volume of water that is evaporated from Earth’s surface is 1.2 x 1012 m3,
Mass of evaporated water, m = r x V
= 1.0 x 103 x 1.2 x 1012
= 1.2 x 1015 kg
Energy required Q = mlv
= 1.2 x 1015 x 2.3 x 106
= 2.8 x 1021 J
2.
A solar furnace made from a concave mirror of area 0.40 m2 is used to heat water. The radiant energy from the Sun arrives at the mirror surface at a rate of 1400 W m-2. The specific heat capacity of water is 4200 J kg-1 K-1. What is the best estimate of the least time that the furnace takes to heat 1.0 kg of water from 20°C to 50°C? [225 s]
Solution
Q = m c Dq
Since , where ΔE is the energy Q required to raise the temperature of H2O,
ΔE = Q = P t
P t = m c Δθ
You will need to know that intensity I of the of radiant energy and area A used is related to power P of the radiant energy from the Sun by the equation
Intensity
P = IA
IAt = mcΔθ
(1400)(0.40) t = (1.0) (4200) (50 – 20)
t = 230 s
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3.
A 1.00 kg glass jug [c = 840 J kg-1 K-1] contains 2.00 kg of beer [c = 4200 J kg-1 K-1] at 28.0 °C. 0.500 kg of ice [c = 2000 J kg-1 K-1] are added to the beer. The ice has an initial temperature of - 5.0 °C. It may be assumed that the jug do not lose any heat to the environment. The latent heat of fusion of ice is 3.35 x 105 Jkg-1. When thermal equilibrium is reached, all the ice has melted and the final temperature of the mixture is above 0 °C. Determine this temperature θ. [7.6 °C]
Solution
When the ice is put into the jug of beer, thermal energy is transferred from the jug and beer to the ice (since there is a temperature difference between the ice and the jug & beer.)
Note: during the heat transfer to the ice, the temperature of the ice
1. increase from – 5.0 °C to 0°C,
2. then constant at 0 °C when the ice melts and
3. increase from 0 °C to T.
Hence these 3 different heating processes has to be taken account when calculating the heat gained by ice.
Magnitude of heat lost by jug and beer = Magnitude of heat gained by ice
Qjug+beer (28 – T) = Qice (– 5.0 °C to 0°C) + Qmelting ice + Qwater (0 °C to T)
Qjug+beer (28 – T)
= m c Dq
= [(1.00 x 840) + (2.00 x 4200)] (28.0 – T)
= 258 720 – 9240 T
Qice (– 5.0 °C to 0°C)
= [0.50 x 2000 x (0 – (-5.0)]
= 5000 J
Qmelting ice
= mlf
= 0.50 x 3.35 x 105
= 167 500 J
Qwater (0 °C to T)
= m c Dq
= 0.50 x 4200 x (T – 0)
= 2100 T
Total heat lost by jug and heat = Total heat gained by ice (from 0 °C to T)
258 720 – 9240 T = 5 000 + 167 500 + 2100 T
11 340 T = 86 220
T = 7.6 °C
Conversion of temperature and Use of Ideal Gas Equation
4.
Some cars are fitted with bags packed into steering column. In an accident, gas is forced under pressure into the bag and the bag of gas quickly acts as a cushion between the driver and steering wheel. In one such system, the volume of gas used in the bag is 0.037 m3 when the pressure is 1.8 x 105 Pa and the temperature is 6.0 °C.
Calculate
(i) the temperature of the gas in kelvins, [279 K]
(ii) the amount of gas used, in moles, [2.87 mol]
(iii) the pressure in the bag when the temperature rises to 18 °C assuming the volume to remain constant while the temperature rises. [1.88 x 105 Pa]
Solution
(i) T / K = T / oC + 273.15
= 6 + 273.15
= 279 K
(ii) pV = nRT
1.8 × 105 × 0.037 = n × 8.31 × 279
n = 2.87 mol
(iii) Assuming the volume to remain constant,
pV = nRT
p × 0.037 = 2.87 × 8.31 × (273.15 + 18)
p = 1.88 × 105 Pa
(Alternatively)
pV = nRT
p = 1.88 x 105 Pa
Internal Energy
5. [J03/4/2 part]
On the following figure, place a tick (√) against those changes where the internal energy of the body is increasing.
Internal energy U is the summation of the kinetic energies Ek and potential energies Ep of the molecules in the system.
Ek / Ep / UWater freezing at constant temperature / - / ↓ / ↓ / ×
A stone falling under gravity in vacuum / - / - / - / ×
Water evaporating at constant temperature / - / ↑ / ↑ / P
Stretching a wire at constant temperature / - / ↑ / ↑ / P
First Law of Thermodynamics
6. [N93/3/5]
Some solids contract and some solids expand when they melt. Complete the table with the symbols + or − to indicate the signs of the thermodynamic quantities for each of the two types of solids when the solids melt at constant pressure.
Solids whichcontract on melting / Solids which
expands on melting
/ + / +
Q / + / +
W / + / -
During melting, the internal energy of the molecules of the solid increases. Basically, the thermal energy supplied is used to weaken the force of attraction between molecules. This increases the distances between molecules, hence increases the potential energy of the molecules while keeping kinetic energy constant (constant temp).
Heat is supplied to the solid during melting process.
The work done on the solid for contraction and expansion are positive and negative respectively.
First Law of Thermodynamics
7. HCJC 2001 Prelim Paper 3
An ideal gas expands from a volume of 2.00 m3 to a volume of 6.00 m3 along two different paths as described in the figure below.
The heat added to the gas along the path I→A→F is equal to 1.67 × 106 J.
(a) State in words the First Law of Thermodynamics.
(b) Compute the increase in the internal energy of the gas as it undergoes expansion process I→A→F. [1.07 x 106 J]
(c) Hence, compute the heat added to the gas along the path I→F. [1.47 x 106 J]
Solutions
(a) The increase in the internal energy of a system is the summation of heat supplied to system and work done on system.
(b) The heat supplied to gas along the path I→A→F is + 1.67 × 106 J.
Thus to find the increase in the internal energy of the gas along I→A→F, the work done on gas along I→A→F has to be found 1st.
Workdone from I→A→F
= area under the graph of I→A→F
= ½ x (6.00 – 2.00) x (2.00 – 1.00) x 105 + (1.00 x 105) x (6.00 – 2.00)
= 6.00 x 105 J
Thus, ΔU = Q + W
ΔUI→A→F = (+ 1.67 x 106) + (- 6.00 x 105)
= + 1.07 x 106 J
(c) The starting and ending point of the two processes (I→A→F and I→F) are the same. Hence the starting and ending temperatures of the gas undergoing the two processes will be the same. Therefore the change in temperature for the gas through the two processes is the same as well.
Since internal energy is dependent on the temperature (), the change in internal energy for the two processes should be the same as well.
ΔUI→F = ΔUI→A→F
= + 1.07 x 106 J
To compute the heat supplied to the gas, the work done on gas through I→F has to be found.
W = (1.00 x 105) x (6.00 – 2.00)
= 4.0 × 105 J
Work done on gas = - 4.0 × 105 J (expansion)
ΔU = Q + W
1.07 x 106 = Q + (- 4.0 × 105)
Q = + 1.47 x 106 J
Use of Ideal Gas Equation
8. J96/3/5 part
(a) The air cylinder for a diver has a volume of 9.00 x 103 cm3 and when the cylinder is filled, the air has a pressure of 2.10 x 107 Pa at 24oC. The diver is swimming in water of density 1.03 x 103 kg m-3 and temperature 24oC at a depth of 15.0 m. When the diver breathes in, the pressure of air delivered from the cylinder to the diver is always equal to the pressure of the surrounding water.
Atmospheric pressure is 1.01 x 105 Pa.
Calculate for the depth of 15.0 m,
(i) the total pressure on the diver [2.53 x 105 Pa]
(ii) the volume of air available at this pressure from the cylinder. [7.47 x 105 cm3]
(b) The supply of air in (a) is sufficient for the diver to remain at a depth of 15.0 m for 45 minutes. Assuming that the diver always breathes at the same rate (i.e. the same volume is required per minute, regardless of pressure), how long would the air in the cylinder last for the diver at a depth of 35.0 m and a water temperature of 20oC. [24.7 mins]
Solutions
(a) (i) pressure due to water, P = h r g
= (15.0) (1.03 x 103) (9.81) = 1.52 × 105 Pa
Total pressure = atmospheric pressure + water pressure
= 1.01 × 105 Pa + 1.52 × 105 Pa
= 2.53 × 105 Pa
(ii) pV = nRT
For the same amount of gas and at constant temperature: n1 = n2 and T1 = T2
V = 7.47 x 105 cm3
(b) Pressure at depth of 35.0 m = 1.01 × 105 Pa + (35.0) (1.03 x 103) (9.81)
= 4.55 × 105 Pa
V’ = 4.10 x 105 cm3.
Since volume of air µ time underwater
(1): 7.47 x 105 cm3 µ 45 mins
(2): 4.10 x 105 cm3 µ T mins
(2)/(1): T = 24.7 mins
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Concept of Mole
9. Gold is traded in the Commodities Market at US$ 900.000 per troy ounce. One troy ounce is exactly 31.1034768 g.
Suppose you are asked to buy 1 mole of gold, how much will it cost you?
How many atoms of gold can you buy with US$ 100 000?
1 mole of gold has a mass of 196.966569 g
[US$ 5699.36, 1.05626 x 1025 atoms]
Solution
US$ 900.000 buys 31.1034768 g
1 g cost US$ 900.000 /31.1034768 g
1 mole cost US$ 900.000 x 196.966569 /31.1034768 = US$ 5699.36
US$ 5699.36 buys 6.02 x 1023 atoms of gold
US$ 100 000 buys 6.02 x 1023 x 100 000 / 5699.36 = 1.05626 x 1025 atoms
Conceptual Questions
10. Define Internal energy and hence comment briefly on this statement “This substance contains heat”.
Solution
Internal energy is determined by the state of the system. Internal energy of a system is the sum of random distribution of KE and PE of all the molecules in the system.
The substance contains internal energy, not heat. The word “heat” is used only when referring to the thermal energy actually in transit from hot to cold.
11. A flask containing boiling water is removed from the burner. A cork is then placed in the neck of the flask to seal it. To restart the boiling, cold water is poured over the neck of the flask. Explain briefly why such an act leads to reboiling.
Solution
Cold water causes some vapour in the flask to condense. Hence the pressure above the liquid in the flask drops. When the pressure drops to a certain level, the amount of energy needed by the water molecules in the flask to escape into the air above is lesser, as such, the water boils.
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12. When a bubble rises in a glass of beer, it is observed that it doubles in volume by the time it reaches the surface. What is the cause of the doubling in volume? Given that you may assume the beer to be of the same temperature throughout the glass and the height of a beer glass is 0.200 m.
Solution
Note: when a bubble rises from the bottom of a beer glass, the pressure experience by this bubbles decreases. Assuming there is no change in the temperature, there will be an increase in the volume of the bubble. However, to have its volume doubled solely due to decreases in pressure, the beer glass of height of 10.3 m will be needed. There is not likely to have such a tall glass.
In fact, the bubble in a beer act as a nucleation site for CO2 molecules, so as the bubble rises, it accumulates carbon dioxide from the surrounding beer and grows larger, implying n is now bigger. Using the equation of state, with constant temperature and with the decrease in pressure as the bubble rises, the bubble doubles in volume.