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Chapter 12:Temperature and Heat ( Physics 2010 - Pellissippi )
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The temperature of an object is
(b) a measure of how hot or cold the object is
(c) a measure of the average K.E. of the atoms and molecules of the object
The regular temperature scales are
(b) Celsius and Fahrenheit
The absolute temperature scales are
(a) Kelvin and Rankin
The reason for using the melting point of a pure substance in the calibration of thermometers is that
(a) during melting, for example, the temperature of a pure substance remains constant
(b) it can be replicated at other similar points on this planet
The reason for using the boiling point of a pure substance in the calibration of thermometers is that
(a) during boiling, for example, the temperature of a pure substance remains constant
(b) it can be replicated at other similar points on this planet
A phase change for a pure substance means going from
(a) solid phase to liquid phase or vice versa
(b) from liquid phase to vapor phase or vice versa
The basis for calibration of the Celsius scale is the melting and freezing points of
(a) water
Fahrenheit scale also uses the melting and boiling points of water as basis for calibration
(b) False.
The zero of Celsius scale corresponds to
(a) 32 F
The 100 on the Celsius scale corresponds to
(c) 212 F
Conversion from C to F is
(b) F = (9/5)C + 32
Conversion from F to C is
(b) C = (5/9)(F-32)
If a temperature difference in Celsius scale is ΔC = 20o, the corresponding ΔF is
(b) 36
If a temperature difference in Fahrenheit scale is ΔF = 72o, the corresponding ΔC is
(a) 40
The temperature at which both scales read the same number is
(c) -40
In Fig. 2, V1 is the volume at T1 =100oC, and V2 the volume at T2 = 50oC. V3 is the volume at T3 = -100oC and V4 is the volume of the gas at T4 = -200oC. The ratio of (V1-V2) / (V3-V4) as is apparent from the line segments is
(c) 1/2
In Fig. 2, the ratio of (T1-T2) / (T3-T4)
(c) 1/2
From the above two questions, one may conclude that
(a) ΔV/ΔT = constant
ΔV/ΔT = constant means that the variations of V with respect to T is
(c) linear
Since V varies linearly with T, it is then (a) incorrect (b) correct (c) partially correct to extrapolate the volume-temperature graph of Fig. 2 like a straight line to cross the temperature axis at -273C.
(b) correct
Conversion from oF to Rankin is
(b) R = oF+460
Heat is a type of energy that flows due to
(b) temperature difference
The SI unit for heat is
(b) Joule
1kcal is the amount of heat that can change the temperature of
(c) 1kg of water by 1 C
1Btu is the amount of heat that can change the temperature of 1 lbm of water by
(a)1 F
1 cal is the amount of heat that can change the temperature of 1gram of
(c) water by 1 C
Specific heat of a substance is the amount of heat that can change the temperature of 1 unit of
(a) mass
It takes (a) more (b) less amount of heat to warm up 1 kg of aluminum than 1 kg of water by 1oC.
(b) less amount
The specific heat of water is
(a) 1cal /(gr oC)
(b) 1kcal /(kg oC)
(c) 1Btu /(lbm oF)
The formula for the amount of heat with no phase change is
(a) Q = McΔT
(b) Q = Mc(Tf -Ti)
The formula that calculates the amount of heat given or taken in a phase change is
(a) Q = MLf
(b) Q = MLV
The amount of heat that warms up 2.5kg of aluminum at 27oC to 67oC is
(a) 21.5kcal
(b) 21500cal
The amount of heat that warms up 4.0kg of aluminum at 27.0oC by 50.0oC is
(a) 43.0kcal
The amount of heat necessary to convert 5.0kg of ice at 0oC to water at 50.0oC is
(b) 650kcal
The amount of heat necessary to convert 5.0kg of ice at - 40.0oC to water at 50.0oC is
(a) 748kcal
The amount of heat necessary to convert 5.0kg of ice at - 40.0oC to boiling water (100oC) is
(b) 996 kcal
The amount of heat that 2.00kg steam at 100oC must lose to become water at 20.0oC is
(b) -1240kcal
When 4.00kg of water at 25.0oC is mixed with 2.00kg of water at 75.0oC in a well-insulated container, the equilibrium temperature is
(a) 41.7 C
When 8.00kg of aluminum at 200C is placed in 3.00kg of water at 25.0o, knowing that heat exchange occurs between water and aluminum only, the equilibrium temperature is
(c) 88.8 C
255gr hot iron is placed in 65.0gr of water at 18.5o. The equilibrium temperature is 88.5oC. The initial temperature of the iron piece is
(a) 254
In general, except for a few substances at certain specific temperatures, all elements expand due to temperature increase.
(a) True
Thermal expansion is due to
(a) increased rate of oscillations of atoms (b) the extra space each atom requires because of increased amplitude of oscillation
For expansion of solids, generally two variables are considered:
(a) length & temperature
For linear expansion of a wire, the change in length ΔL is proportional to
(a) the length of the wire L
(b) just the change in temperature ΔT
Based on the previous question, we may write: (a) ΔL = αL (b) ΔL= αLΔT (c) ΔL = α ΔT where α is the proportionality factor called the "Coefficient of Linear Thermal Expansion."
solving for α gives: (a) α = ΔL/L (b) α = ΔL/(LΔT) (c) α = ΔL/(ΔT)
(b) ΔL= αLΔT
(b) α = ΔL/(LΔT)
Change in length per unit length is
(c) ΔL/L
(Change in length per unit length) per unit change in temperature, may be written as
(c) ΔL/L/ΔT
ΔL/L/ΔT is the same thing as ΔL/(LΔT).
(a) True
The coefficient of Linear Expansion α is defined as the "change in length per unit length per unit change in temperature."
(a) True
If we write ΔL as Lf - Li , α becomes α = (Lf - Li)/(L ΔT). Solving for Lf yields
(a) Lf = Li + LiαΔT
(b) Lf = Li [1+αΔT]
The term Liα ΔT is therefore
(a) the overall change in length
(b) the change in the initial length
The reason we may simply use 2α for the area expansion of a solid sheet is that, in the derivation, the term ( ΔL)2 is
(a) very small compared to other terms
The reason we use 3α for volume expansion of a solid is that, in the derivation, the terms (ΔL)2 and (ΔL)3 are
(a) very small compared to other terms
For expansion of liquids, there is only one formula as opposed to the expansion of solids that have 3 formulas.
(a) True
The unit for α = ΔL/(LΔT) is
(a) (temperature)-1 (b) C-1
(c) F-1
To calculate ΔL for a copper power line 500 km long for a temperature change of ΔT =150oC, the equation ΔL = Liα ΔT may be used. The result is
(a) 1.28km
The formula that calculates the change in the volume of a liquid is
(b) ΔV = Viβ ΔT
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