1. The equivalent of a glass of water (that is, 270 g of liquid) at 20.0 °C receives 1000 cal from a candle flame. Assuming that all the energy goes into the water without any losses, what is the final temperature of the liquid?
SOLUTION
Q = 1000 cal = cm( - )
Q = (1.00 cal/g∙C˚)(270 g)( - 20.0 )
and
= (1000) /((1.00 cal/g∙C˚)(270 g)) +20.0˚C = 23.7
2. A cooper pot has a mass of 0.50 kg and is at 100. How much heat must be removed from it if its temperature is to be lowered to precisely 0?
SOLUTION
Given : cu, m = 0.50 kg,
= 100
= 0
∆T = -100 K
Q = m∆T
= (390 J/kg∙K)(0.50 kg)(-100 K)
= -19.5kJ / -20kJ
3. A ontainer holding 0.250 kg of water at 20.0 is placed in the freezer compartment of refrigerator. How much energy must be removed from the water to turn it into ice at 0 ?
SOLUTION
Given : water, = 0.250 kg
= 20.0
=0
Q = ( - ) + (-)
Q = (4.186 kJ / kg∙K)(0.250 kg)(0 - 20.0 - (0.25kg)(334 kJ / kg)
= -20.9 kJ – 83.5kJ
= -1.04 105 J
4. How much heat must be added to a 1.0 kg mass to of water ice at -10 ˚C and atmospheric pressure, in order to transform into superheated steam at 110 ˚C?
SOLUTION
Given : m = 1.0 kg
= -10 ˚C
= 110 ˚C
Q = ∆ + + ∆ + + ∆
Q = (1.0 kg)(2.1 kJ/kg∙K)[0 ˚C –(-10˚C)] + (1.0 kg)(334 kJ/kg)
+(1.0 kg)(4.2 kJ/kg∙K)[100˚C-(0˚C)] + (1.0 kg)(2.26 103 kJ/kg)
+(1.0 kg)(2.0 kJ/kg∙K)[110˚C-100˚C
Q = 21 kJ + 334kJ + 420kJ + 2260 kJ + 20kJ
= 3055kJ
5. A slab having thickness of 4cm and measuring 25cm on the side has a 40 ͦ C temperature difference between its faces. How much heat flows through it per hour? the conductivity k is 0.0025 cal/s . cm . ͦ C.
Q =[k (area)(th – tc)(time)] / thickness = 0.0025(25 x 25)(40)(3600 / 4
= 56.25 kcal.
6. A refrigerator door is 150cm high, 80cm wide, and 6cm thick. If the coefficient of conductivity is 0.0005 cal/cm . s . ͦ C, and the inner and outer surfaces are at 0 and 30 ͦ C, respectively, what is the heat loss per minute through the door, in calories?
Q = kA(th – tc)(time) / d = 0.0005(150 x 80)(30 ͦ C - 0 ͦ C)(60) / 6
= 1800cal.
7. how much water at 100 ͦ C could be evaporated per hour by the heat transmitted through a 1cm x 1cm steel plate 0.2cm thick, if the temperature difference between the plate faces is 100 ͦ C? For steel, k is 0.11cal/s . cm . ͦ C.
Q = (0.11cal/s . cm . ͦ C)(1cm2)(100 ͦ C)(3600s)
0.2 cm
= 198 kcal.
8. What thickness of wood has same insulating ability as 8cm of brick?
k = 0.8W/m . K for brick and 0.1W/m .K for wood.
:: for the same ∆T and A, the ∆Q / ∆t is the same in two materials, so
Kw / Lw = kb /Lb, and the wood thickness is Lw = (0.1/0.8)(8cm)
= 1cm.
9. Distinguish between natural and forced convection.
• If the motion of a fluid is caused by a difference in density that accompanies a change in temperature, the current produced is referred to as natural convection. When a fluid is caused to move by the action of a pump or fan, the current produced is referred to as forced convection. Both type of convection are employed in a common home heating system : hot water is circulated through “radiators” by forced convection; the warmed air rise by natural convection.
10. Forced air flows over a convection heat exchanger in a room heater, resulting in a convective heat-transfer coefficient h = 200Btu/h . ft^2 . ͦF. the surface temperature of the heat exchanger may be considered constant at 150 ͦF, and the air is at 65 ͦF. determine the heat exchanger surface area required for
30 000Btu/h of heating.
H= hA∆T. from the information given,
(30 000Btu/h) = (200Btu/h . ft^2 . ͦF)A A = 1.765 ft^2.
Sunday, February 28, 2010
question for temperature and heat
1. the ampification or gain of a transistor amplifier may depend on the temperature. The gain for a certain amplifier at room temperature (120.0○C) is 30.0, where as at 55○C it is 35.2. what would the gain be at 28○C if gain depend linearly on temperature over this limited range?
35.2 – 30.0 = 5.2
55.0 – 20.0 = 35.0○C
5.2 \ 35.0 = 0.15
28.0 -20.0 = 8.0
0.15 × 8.0 = 1.2
30.0 +1.2 = 3.2
2. A friend suffering from the flu has a fever; her body temperature is 38.6○C. What is her temperature in Kelvin?
Tk = 38.6 + 273.15 = 311.8 K
3. A cylinder contains 250L of xenon gas (Xc) at 20.0○C and pressure af 5.0 atm. How much heat is required to raise the temperature of this gas to 50.0○C, holding the volume constant? Threat the xenon aas an ideal gas.
P = 5.0 atm = 5.0 × (1.01×105 pa)
V = 250L = 250 × 10-3m3
T = 20.0○C + 273.15 = 293.15 ○K
N = PV/RT = (5.05×105pa) × (250×10-3m3)/(8.31J/mol.k)(393.15K) = 51.8 mol
∆T = 50.0 -20.0 = 30.0○C
Q = (3/2) n R∆T = (3/2)(51.8 mol)(8.31 J/ mol.C)(30.0○C) = 19kJ
4. A jewelry designer plans to make some specially ordere of silver charms for a commemorative braclet. If the melting point of silver is 960.8○C, how much heat must jeweler add to 0.500kg of silver at 20.0○C to be able to silver into her charm molds?
Q = mc∆T + mLf
Q= (0.500Kg)(0.235kJ/kg.C)(960.8○C-20.0○C) + (0.500Kg)(88.3kJ/kg)
= 110.5kJ + 44.15 kJ = 155kJ
5. A healthy person has an oral temperature of 98.6○F. what would this reading be an this Celsius scale.
F= 98.6○F - 32○F = 66.6○F
C=66.6○F(1/(9/5)) = 37.0○C
6. A block of concrete has a cross section of 8 m and a thickness of 10 cm. One side is at 40 C, and other is at 20 C. What is the rate of heat transfer?
Q/t = k . A [ (Thot – T cold) /d ]
= ( 1.7 Wm-1 K-1 ). (8m2). [(40○C – 20○C) /0.10 ]
= 2720W
7. Your whole family has taken showers before your, dropping the temperature in the water heater to 18○C. If the heater holds 150kg of water, how much energy will it take to bring it to 50○C?
∆Q = mc ∆T = (150Kg)(418.4J/kg.K)(50○C-18○C)
= 2.0 × 107 J
8. What temperature gradient must exist in a copper rod for it to transmit 33.5W/m2 of cross 10 ×10-5m down the rod?
Q/t = k. A [ (Thot – T cold) /d ]
33.5W/m2 = (385 Wm-1 K-1 ).( 10 ×10-5m) [ (Thot – T cold) /d ]
[ (Thot – T cold) /d ] = 33.5W/m2 ÷ (385 Wm-1 K-1 ).( 10 ×10-5m)
= 870○C/m
9. A steel beam is used in the roadbed of a brelge. The beam is mounted between two concrete support when the temperature is 23○C, with no room provided for thermal expansion. What compressional stress mush the concrete support apply to each end of the beam, if they are to keep the beam from expanding when the temparute rises to 42○C?
Stress= у∞∆T = (2.0×1011N/m2)(12×10-6(C)-1)(42○C-23○C) = 4.6 ×107N/m2
10. How much heat is required to raise the temperature of 0.20kg of water from 15○C to 45○C?
Q = mc∆T = (0.20kg)(4186J/kg.C)(45○C-30○C) = 2.5 × 104J
35.2 – 30.0 = 5.2
55.0 – 20.0 = 35.0○C
5.2 \ 35.0 = 0.15
28.0 -20.0 = 8.0
0.15 × 8.0 = 1.2
30.0 +1.2 = 3.2
2. A friend suffering from the flu has a fever; her body temperature is 38.6○C. What is her temperature in Kelvin?
Tk = 38.6 + 273.15 = 311.8 K
3. A cylinder contains 250L of xenon gas (Xc) at 20.0○C and pressure af 5.0 atm. How much heat is required to raise the temperature of this gas to 50.0○C, holding the volume constant? Threat the xenon aas an ideal gas.
P = 5.0 atm = 5.0 × (1.01×105 pa)
V = 250L = 250 × 10-3m3
T = 20.0○C + 273.15 = 293.15 ○K
N = PV/RT = (5.05×105pa) × (250×10-3m3)/(8.31J/mol.k)(393.15K) = 51.8 mol
∆T = 50.0 -20.0 = 30.0○C
Q = (3/2) n R∆T = (3/2)(51.8 mol)(8.31 J/ mol.C)(30.0○C) = 19kJ
4. A jewelry designer plans to make some specially ordere of silver charms for a commemorative braclet. If the melting point of silver is 960.8○C, how much heat must jeweler add to 0.500kg of silver at 20.0○C to be able to silver into her charm molds?
Q = mc∆T + mLf
Q= (0.500Kg)(0.235kJ/kg.C)(960.8○C-20.0○C) + (0.500Kg)(88.3kJ/kg)
= 110.5kJ + 44.15 kJ = 155kJ
5. A healthy person has an oral temperature of 98.6○F. what would this reading be an this Celsius scale.
F= 98.6○F - 32○F = 66.6○F
C=66.6○F(1/(9/5)) = 37.0○C
6. A block of concrete has a cross section of 8 m and a thickness of 10 cm. One side is at 40 C, and other is at 20 C. What is the rate of heat transfer?
Q/t = k . A [ (Thot – T cold) /d ]
= ( 1.7 Wm-1 K-1 ). (8m2). [(40○C – 20○C) /0.10 ]
= 2720W
7. Your whole family has taken showers before your, dropping the temperature in the water heater to 18○C. If the heater holds 150kg of water, how much energy will it take to bring it to 50○C?
∆Q = mc ∆T = (150Kg)(418.4J/kg.K)(50○C-18○C)
= 2.0 × 107 J
8. What temperature gradient must exist in a copper rod for it to transmit 33.5W/m2 of cross 10 ×10-5m down the rod?
Q/t = k. A [ (Thot – T cold) /d ]
33.5W/m2 = (385 Wm-1 K-1 ).( 10 ×10-5m) [ (Thot – T cold) /d ]
[ (Thot – T cold) /d ] = 33.5W/m2 ÷ (385 Wm-1 K-1 ).( 10 ×10-5m)
= 870○C/m
9. A steel beam is used in the roadbed of a brelge. The beam is mounted between two concrete support when the temperature is 23○C, with no room provided for thermal expansion. What compressional stress mush the concrete support apply to each end of the beam, if they are to keep the beam from expanding when the temparute rises to 42○C?
Stress= у∞∆T = (2.0×1011N/m2)(12×10-6(C)-1)(42○C-23○C) = 4.6 ×107N/m2
10. How much heat is required to raise the temperature of 0.20kg of water from 15○C to 45○C?
Q = mc∆T = (0.20kg)(4186J/kg.C)(45○C-30○C) = 2.5 × 104J
Question Heat and Temperature...
1. How many kilocalories of heat must be added to 10.0 kg of steel to raise its temperature 150°C?
Solution :
Data :
m = 10.0kg
ΔT = 150°C
c = 0.115kcal/kg °C
Q = ?
Q = vmΔT
Q = [(0.115)/(kcal/kg°C)](10.0kg)(150°C)
=173kcal
2. What is the specific heat of a substance that absorbs 2.5 x 103 joules of heat when a sample of 1.0 x 104 g of the substance increases in temperature from 10.0C to 70.0C?
Solution :
Q= (m)(Dt)(cp) where q is the energy in joules m is the mass Dt is the change in initial temperature to the final temperature cp is the specific heat (heat required to raise 1 g of the substance 1C)
Q= (m)(Dt)(cp)
2.5 x 103 J = (1.0 x 104 g)(70C-10C)(cp)
[2.5 x 103 J] = cp
[1.0 x 104 g][70C-10C]
0.0042 J = cp
gC
cp = 0.0042J/gC
3. A 1.0 kg sample of metal with a specific heat of 0.50 KJ/KgC is heated to 100.0C and then placed in a 50.0 g sample of water at 20.0C. What is the final temperature of the metal and the water?
Solution :
Qmetal = (m)(Dt)(cp) Qwater = (m)(Dt)(cp)
Qmetal = qwater
(m)(Dt)(cp) = (m)(Dt)(cp)
[1.0Kg][100C-Tf][0.50KJ] = [50.0g][Tf-20C][4.184 J]
[ KgC] [ gC ]
[1.0Kg][1000g][100C-Tf][0.50KJ][ 1Kg][1000J] = [50.0g][Tf-20C][4.184J]
[ 1Kg] [ KgC][1000g][ 1KJ] [ gC ]
[1000g][100C-Tf][0.50J] = [50.0g][Tf-20C][4.184 J]
[ gC] [ gC ]
[1000g][100C-Tf][0.50 J/gC] = [Tf-20C]
[ 50.0g] [4.184 J/gC]
[2.39][100C-Tf] = [Tf-20C]
239C-2.39Tf = Tf-20C
239C + 20C = Tf + 2.39Tf
259C = 3.39Tf
259C = Tf
3.39
76C = Tf
4. A 2.8 kg sample of a metal with a specific heat of 0.43KJ/KgC is heated to 100.0C then placed in a 50.0 g sample of water at 30.0C. What is the final temperature of the metal and the water?
Solution :
Qmetal = (m)(Dt)(cp) Qwater = (m)(Dt)(cp)
Qmetal = Qwater
(m)(Dt)(cp) = (m)(Dt)(cp)
[2.80Kg][100C-Tf][0.43KJ] = [50.0g][Tf-30C][4.184 J]
[ KgC] [ gC ]
[2.8.0Kg][1000g][100C-Tf][0.43KJ][ 1Kg][1000J] = [50.0g][Tf-30C][4.184 J]
[ 1Kg] [ KgC][1000g][ 1KJ] [ gC ]
[2800g][100C-Tf][0.43J] = [50.0g][Tf-30C][4.184 J]
[ gC] [ gC ]
[2800g][100C-Tf][0.43 J/gC] = [Tf-30C]
[ 50.0g] [4.184 J/gC]
[24.08][100C-Tf] = [Tf-30C]
2408C-24.08Tf = Tf-30C
2408C + 30C = Tf + 24.08Tf
2438C = 25.08Tf
2438C = Tf
325.08
97C = Tf
5. How much heat is needed to transform a 1 kg block of ice at –5ºC to a puddle of water at 10ºC?
Solution :
Q = mcΔT
= (1kg)(2.20 x 10³ J/kg°C)(5°C)
= 1.1 x 10^4 J
Q = mfusion
= (1kg)(3.3 x 10^5J/kg)
= 3.3 x 10^5 J/kg
1.1 x 10^4 + 3.3 x 10^5 + 4.2 x 10 ^4 = 3.8 x 10⁶
6. 4190 J of heat are added to 0.5 kg of water with an initial temperature of 12ºC. What is the temperature of the water after it has been heated?
Solution :
ΔT = Q/mc
= (4190 J)/(0.5kg x 4190 J/kg °C)
= 2 °C
7. A 260 g steak is resting on a 200 g glass plate. The initial temperature of the steak is 85 ° C and the initial temperature of the plate is 30 ° C. The heat capacity of the steak is approximately 0.5 J/kg K and the heat capacity of the glass plate is 1.2 J/kg K. Determine the final temperature of the steak/plate system.
Solution :
ms = 260 g
mp = 200 g
Cps = 0.5 J/kg K
Cpp = 1.2 J/kg K
T1,s = 85 ° C
T1,p = 30 ° C
TF = [-(0.260 kg)*(0.5J/kg K)*(358 K) – (0.200 kg)*(1.2J/kg K)*(303 K)]
[-(0.260 kg)*(0.5J/kg K) – (0.200 kg)*(1.2J/kg K)]
TF = 322 K
8. Determine the heat transferred between a 260 g steak resting on a 200 g plate. The thermal conductivity of the steak is 0.49 W/m K and the area the steak covers on the plate is 10 cm2. The thickness of the steak is 3.5 cm. Again the initial temperature of the steak and plate are 85 ° C and 30 ° C respectively.
Solution :
Q = -kA(D T)/( D x)
Q = heat conducted
k = thermal conductivity
A = surface area for conduction
D T = TF – Ti
D x = thickness
k = 0.49 W/m K
A = 10 cm2
D T = TF – Ti = 358 K – 322 K = 36 K
D x = 3.5 cm
A = 0.001 m2
D x = 0.035 m
Q = -[(0.49W/m K)*(0.001 m2)*(36 K)]/(0.035 m)
Q = -0.504 WF
9. A 10.0-lb pieace of hot copper is dropped into 30.0lb 0f water at 50°F. If the final temperature of the mixture is 65°F, what was the initial temperature of the copper?
Solution :
w1 = 10.0lb wg = 30,0lb
c1 = 0.093Btu/lb °F cg = 1.00Btu/lb °F
T1 = ? Tg = 50 °F
c1w1 (T1-Tf) = cgwg (Tf-Tg)
T1 = (cgwg / c1w1)(Tf-Tg) + Tf
T1 =[ (1.00Btu/lb °F )(30.0lb) / (0.093Btu/lb °F)(10.0lb)] (65 °F-50 °F) + 65 °F
= 550 °F
(0.093Btu/lb °F)(10.0lb)( T1- 65 °F) = (1.00Btu/lb °F)(30.0lb)(65 °F-50°F)
0.93T1Btu/°F-60 Btu = 450 Btu
0.93T1Btu/°F =510 Btu
Tl = (510Btu) / (0.93Btu/ °F)
T1= 550 °F
10. A steel railroad rail is 40.0 ft long at 0°F. How much will it expand when heated to 100 °F?
Solution :
l = 40.0 ft ΔT = 100 °F - 0° F = 100 °F
α = 6.5 x 10ˉ⁶ / °F Δl = ?
Δl = αlΔT
Δl = (6.5 x 10 ˉ⁶/°F)(40.0ft)(100°F)
= 0.026ft or 0.31in
1. How many kilocalories of heat must be added to 10.0 kg of steel to raise its temperature 150°C?
Solution :
Data :
m = 10.0kg
ΔT = 150°C
c = 0.115kcal/kg °C
Q = ?
Q = vmΔT
Q = [(0.115)/(kcal/kg°C)](10.0kg)(150°C)
=173kcal
2. What is the specific heat of a substance that absorbs 2.5 x 103 joules of heat when a sample of 1.0 x 104 g of the substance increases in temperature from 10.0C to 70.0C?
Solution :
Q= (m)(Dt)(cp) where q is the energy in joules m is the mass Dt is the change in initial temperature to the final temperature cp is the specific heat (heat required to raise 1 g of the substance 1C)
Q= (m)(Dt)(cp)
2.5 x 103 J = (1.0 x 104 g)(70C-10C)(cp)
[2.5 x 103 J] = cp
[1.0 x 104 g][70C-10C]
0.0042 J = cp
gC
cp = 0.0042J/gC
3. A 1.0 kg sample of metal with a specific heat of 0.50 KJ/KgC is heated to 100.0C and then placed in a 50.0 g sample of water at 20.0C. What is the final temperature of the metal and the water?
Solution :
Qmetal = (m)(Dt)(cp) Qwater = (m)(Dt)(cp)
Qmetal = qwater
(m)(Dt)(cp) = (m)(Dt)(cp)
[1.0Kg][100C-Tf][0.50KJ] = [50.0g][Tf-20C][4.184 J]
[ KgC] [ gC ]
[1.0Kg][1000g][100C-Tf][0.50KJ][ 1Kg][1000J] = [50.0g][Tf-20C][4.184J]
[ 1Kg] [ KgC][1000g][ 1KJ] [ gC ]
[1000g][100C-Tf][0.50J] = [50.0g][Tf-20C][4.184 J]
[ gC] [ gC ]
[1000g][100C-Tf][0.50 J/gC] = [Tf-20C]
[ 50.0g] [4.184 J/gC]
[2.39][100C-Tf] = [Tf-20C]
239C-2.39Tf = Tf-20C
239C + 20C = Tf + 2.39Tf
259C = 3.39Tf
259C = Tf
3.39
76C = Tf
4. A 2.8 kg sample of a metal with a specific heat of 0.43KJ/KgC is heated to 100.0C then placed in a 50.0 g sample of water at 30.0C. What is the final temperature of the metal and the water?
Solution :
Qmetal = (m)(Dt)(cp) Qwater = (m)(Dt)(cp)
Qmetal = Qwater
(m)(Dt)(cp) = (m)(Dt)(cp)
[2.80Kg][100C-Tf][0.43KJ] = [50.0g][Tf-30C][4.184 J]
[ KgC] [ gC ]
[2.8.0Kg][1000g][100C-Tf][0.43KJ][ 1Kg][1000J] = [50.0g][Tf-30C][4.184 J]
[ 1Kg] [ KgC][1000g][ 1KJ] [ gC ]
[2800g][100C-Tf][0.43J] = [50.0g][Tf-30C][4.184 J]
[ gC] [ gC ]
[2800g][100C-Tf][0.43 J/gC] = [Tf-30C]
[ 50.0g] [4.184 J/gC]
[24.08][100C-Tf] = [Tf-30C]
2408C-24.08Tf = Tf-30C
2408C + 30C = Tf + 24.08Tf
2438C = 25.08Tf
2438C = Tf
325.08
97C = Tf
5. How much heat is needed to transform a 1 kg block of ice at –5ºC to a puddle of water at 10ºC?
Solution :
Q = mcΔT
= (1kg)(2.20 x 10³ J/kg°C)(5°C)
= 1.1 x 10^4 J
Q = mfusion
= (1kg)(3.3 x 10^5J/kg)
= 3.3 x 10^5 J/kg
1.1 x 10^4 + 3.3 x 10^5 + 4.2 x 10 ^4 = 3.8 x 10⁶
6. 4190 J of heat are added to 0.5 kg of water with an initial temperature of 12ºC. What is the temperature of the water after it has been heated?
Solution :
ΔT = Q/mc
= (4190 J)/(0.5kg x 4190 J/kg °C)
= 2 °C
7. A 260 g steak is resting on a 200 g glass plate. The initial temperature of the steak is 85 ° C and the initial temperature of the plate is 30 ° C. The heat capacity of the steak is approximately 0.5 J/kg K and the heat capacity of the glass plate is 1.2 J/kg K. Determine the final temperature of the steak/plate system.
Solution :
ms = 260 g
mp = 200 g
Cps = 0.5 J/kg K
Cpp = 1.2 J/kg K
T1,s = 85 ° C
T1,p = 30 ° C
TF = [-(0.260 kg)*(0.5J/kg K)*(358 K) – (0.200 kg)*(1.2J/kg K)*(303 K)]
[-(0.260 kg)*(0.5J/kg K) – (0.200 kg)*(1.2J/kg K)]
TF = 322 K
8. Determine the heat transferred between a 260 g steak resting on a 200 g plate. The thermal conductivity of the steak is 0.49 W/m K and the area the steak covers on the plate is 10 cm2. The thickness of the steak is 3.5 cm. Again the initial temperature of the steak and plate are 85 ° C and 30 ° C respectively.
Solution :
Q = -kA(D T)/( D x)
Q = heat conducted
k = thermal conductivity
A = surface area for conduction
D T = TF – Ti
D x = thickness
k = 0.49 W/m K
A = 10 cm2
D T = TF – Ti = 358 K – 322 K = 36 K
D x = 3.5 cm
A = 0.001 m2
D x = 0.035 m
Q = -[(0.49W/m K)*(0.001 m2)*(36 K)]/(0.035 m)
Q = -0.504 WF
9. A 10.0-lb pieace of hot copper is dropped into 30.0lb 0f water at 50°F. If the final temperature of the mixture is 65°F, what was the initial temperature of the copper?
Solution :
w1 = 10.0lb wg = 30,0lb
c1 = 0.093Btu/lb °F cg = 1.00Btu/lb °F
T1 = ? Tg = 50 °F
c1w1 (T1-Tf) = cgwg (Tf-Tg)
T1 = (cgwg / c1w1)(Tf-Tg) + Tf
T1 =[ (1.00Btu/lb °F )(30.0lb) / (0.093Btu/lb °F)(10.0lb)] (65 °F-50 °F) + 65 °F
= 550 °F
(0.093Btu/lb °F)(10.0lb)( T1- 65 °F) = (1.00Btu/lb °F)(30.0lb)(65 °F-50°F)
0.93T1Btu/°F-60 Btu = 450 Btu
0.93T1Btu/°F =510 Btu
Tl = (510Btu) / (0.93Btu/ °F)
T1= 550 °F
10. A steel railroad rail is 40.0 ft long at 0°F. How much will it expand when heated to 100 °F?
Solution :
l = 40.0 ft ΔT = 100 °F - 0° F = 100 °F
α = 6.5 x 10ˉ⁶ / °F Δl = ?
Δl = αlΔT
Δl = (6.5 x 10 ˉ⁶/°F)(40.0ft)(100°F)
= 0.026ft or 0.31in
question for temperature and heat
• Change each of the following to the Celsius and Kelvin scales: 68 ͦ F, 5 ͦ F ,176 ͦ F
Tc = ³/9 (tF – 32); Tk = tc + 273.15 = tc + 273. then
tF = 68 ͦ F = tc = 20 ͦ C = Tk = 293 K
tF = 5 ͦF = tc = -15 ͦ C = Tk = 258 K
tF = 176 ͦF= tc = 80 ͦ C = Tk = 352 K
• Hydrogen may be liquefied at – 235 ͦ C under a pressure of 20 atm. What is the temperature on the Fahrenheit scale?
tF = 1.8tc + 32 = 1.8 (-235) + 32 = -391 ͦ F
• The temperature of liquid hydrogen is 20 K. what is the temperature on the Fahrenheit scale?
20 K = 20 – 273 = -253 ͦ C
tF = 1.8tc + 32 = 1.8(-253) + 32 = - 423 ͦF
• Compute the increase in length of 50m of cooper wire when its temperature changes from 12 to 32 ͦC.
∆L = αLo ∆T = (1.67 x 10-³ ͦ C-1)(50m)(20 ͦ C) = 16.7mm.
• A rod 3m long is found to have expanded 0.091cm length for a temperature rise of 60 ͦC. what is α for the material of the rod?
α= 1 /Lo = ∆L /∆T = 0.091 x 10^ -2 m / (3m)(60K) = 5.1 x 10 -6 K^-1
• A brass rod is 0.70m long at 40 o C. fined the length of this rod at 50 ͦ C.
L = Lo (1 + α∆T) = (0.70m){1 + (19.3 x 10^ -6 / ͦ C)}. L = 0.70013 m.
• A block of concrete has a cross section of 10 m2 and the thickness of 0.10m. one side is at 40 ͦ C, and the order is at 20 ͦ C. what is the rate of heat transfer?
Q = k . A (Thot – Tcold / d)
= (1.7 Wm-1K^-1) . (10m^2) . (40 ͦ C – 20 ͦ C / 0.10m)
= 3400 W.
• How much heat is required to raise the temperature of 400kg of mercury from
40 ͦ C to 100 ͦ C?
Q = m.c.∆T = m.c.(Tfinal – Tinitial)
= (0.4kg)(140J/kg ͦ C)(100-40) ͦ C = 3360 J.
• Deep bore holes into the earth show that the temperature increases about 1 ͦ C for each 30m of depth. If the earth’s crust has a thermal conductivity of about 0.80 W/ ͦ C .m, how much heat flows out through the surface of the earth second for each square meter of surface area?
:: One has ∆Q / ∆T = kA (∆T / ∆x).
then ∆Q / ∆t = (0.80W/m . k)(1m2)(1 K/30m)= 27 mW/m2.
• A very small hole in an electric furnace used treating metals acts nearly as a blackbody. If the hole has an area 100mm2, and it is desired to maintain the metal at 1100 ͦ C, how much power travels through the hole?
Power = σAT4 = (3.67 x 10^-8)(10^-4)(1373.15)4 = 20.2 W.
Tc = ³/9 (tF – 32); Tk = tc + 273.15 = tc + 273. then
tF = 68 ͦ F = tc = 20 ͦ C = Tk = 293 K
tF = 5 ͦF = tc = -15 ͦ C = Tk = 258 K
tF = 176 ͦF= tc = 80 ͦ C = Tk = 352 K
• Hydrogen may be liquefied at – 235 ͦ C under a pressure of 20 atm. What is the temperature on the Fahrenheit scale?
tF = 1.8tc + 32 = 1.8 (-235) + 32 = -391 ͦ F
• The temperature of liquid hydrogen is 20 K. what is the temperature on the Fahrenheit scale?
20 K = 20 – 273 = -253 ͦ C
tF = 1.8tc + 32 = 1.8(-253) + 32 = - 423 ͦF
• Compute the increase in length of 50m of cooper wire when its temperature changes from 12 to 32 ͦC.
∆L = αLo ∆T = (1.67 x 10-³ ͦ C-1)(50m)(20 ͦ C) = 16.7mm.
• A rod 3m long is found to have expanded 0.091cm length for a temperature rise of 60 ͦC. what is α for the material of the rod?
α= 1 /Lo = ∆L /∆T = 0.091 x 10^ -2 m / (3m)(60K) = 5.1 x 10 -6 K^-1
• A brass rod is 0.70m long at 40 o C. fined the length of this rod at 50 ͦ C.
L = Lo (1 + α∆T) = (0.70m){1 + (19.3 x 10^ -6 / ͦ C)}. L = 0.70013 m.
• A block of concrete has a cross section of 10 m2 and the thickness of 0.10m. one side is at 40 ͦ C, and the order is at 20 ͦ C. what is the rate of heat transfer?
Q = k . A (Thot – Tcold / d)
= (1.7 Wm-1K^-1) . (10m^2) . (40 ͦ C – 20 ͦ C / 0.10m)
= 3400 W.
• How much heat is required to raise the temperature of 400kg of mercury from
40 ͦ C to 100 ͦ C?
Q = m.c.∆T = m.c.(Tfinal – Tinitial)
= (0.4kg)(140J/kg ͦ C)(100-40) ͦ C = 3360 J.
• Deep bore holes into the earth show that the temperature increases about 1 ͦ C for each 30m of depth. If the earth’s crust has a thermal conductivity of about 0.80 W/ ͦ C .m, how much heat flows out through the surface of the earth second for each square meter of surface area?
:: One has ∆Q / ∆T = kA (∆T / ∆x).
then ∆Q / ∆t = (0.80W/m . k)(1m2)(1 K/30m)= 27 mW/m2.
• A very small hole in an electric furnace used treating metals acts nearly as a blackbody. If the hole has an area 100mm2, and it is desired to maintain the metal at 1100 ͦ C, how much power travels through the hole?
Power = σAT4 = (3.67 x 10^-8)(10^-4)(1373.15)4 = 20.2 W.
question for temperature and heat
Question 1:
85.0 mL of H2O is initially at room temperature (22C). A chilled steel rod at 2C is placed in the water. If the final temperature of the system is 21.2C, what is the mass of the steel bar?
Specific heat of water = 4.18 Specific heat of steel = 0.452
Answer:
THis is a conservation of energy problem; to solve add up thermal energies before and after.
85 x 4.184 x 22 + MS x .452 x 2 = 85 x 4.184 x 21.2 + MS x .452 x 21.2
MS = mass of steel
Solve for MS.
Or you could say that the energy gained by the steel = the energy given up by the water
MS x .452 x [21.2 -2] = 85 x 4.184 x [22 - 21.2]
Question 2
2 HCl(aq) + Ba(OH)2(aq) → BaCl2(aq) + 2 H2O(l) ΔH = −118 kJ
Calculate the heat evolved when 96.4 mL of 0.500 M HCl is mixed with 300.0 mL of 0.445 M Ba(OH)2.
1 kJ
Assuming that the temperature of both solutions was initially 25.0°C and that the final mixture has a mass of 396.4 g and a specific heat capacity of 4.18 J °C-1 g-1, calculate the final temperature of the mixture.
Answer :
96.4 mL of 0.500 M HCl = 0.0482 moles HCl
300.0 mL of 0.445 M Ba(OH)2. = 0.1335 moles Ba(OH)2
by the reaction
2 HCl(aq) + Ba(OH)2(aq) → BaCl2(aq) + 2 H2O(l)
they needed twice as many moles of HCl as there were moles of Ba(OH)2
they did not add enough HCl
HCl is the limiting reagent
find heat:
0.0482 moles HCl @ 118 kJ/2 moles HCl = 2.844 kiloJoules
find dT
dH = mCdT
2,844 joules = 396.4g (4.18 J/g-C)dT
dT = 1.72 hotter
the final temp is 26.7Celsius
Question 3 :
A 0.040 kg ice cube at 0 degrees C is placed in an insulate box that contains 0.0075 kg of steam at 100 degrees C. When the thermal equilibrium is established, all of the ice has melted. Find the final temperature of the closed system. The specific heat of water is 4186 J/kg*K. The latent heat of fusion for ice is 3.34X10^5 J/kg. The latent heat of vaporization for water is 22.6X10^5 J/kg.
Choices:
*22.7 C
*33.6 C
*44.9 C
*50.7 C
*66.4 C
Answer :
Heat given off by steam = Heat absorbed by ice
Qs = Qi
Qs = 0.0075(22.6 x 10^5) + 0.0075(4186)(100 - T)
Qi = 0.040(3.34 x 10^5) + 0.040(4186)(T - 0)
where
T = equilibrium temperature of the system
Since Qs = Qi,
0.0075(22.6 x 10^5) + 0.0075(4186)(100 - T) = 0.040(3.34 x 10^5) + 0.040(4186)(T - 0)
0.1695 x 10^5 + 31.395(100 - T) = 0.1336 x 10^5 + 167.44(T)
Simplifying,
0.1695 x 10^5 + 3139.5 - 31.395T = 0.1336 x 10^5 + 167.44T
Combining terms,
167.44T + 31.395T = 0.1695 x 10^5 - 0.1336 x 10^5 + 0.031395 x 10^5
198.835T = 6729.5
and solving for T,
T = 33.8 C
Question 4:
A 19.0 g sample of ice at -17.0°C is mixed with 102.0 g of water at 86.0°C. Calculate the final temperature of the mixture assuming no heat loss to the surroundings. The heat capacities of H2O(s) and H2O(l) are 2.08 and 4.18 J/g°C, respectively, and the enthalpy of fusion for ice is 6.02 kJ/mol.
Answer :
19.0 g ice at -17.0 deg going to 0 deg requires-
19.0g x 2.08 J/g deg x 17 deg = 671.8 J
Melting 19.0 g ice at 0 deg requires-
19.0g x (1/18.02 g/mol) x 6.02kJ/mol x 1000 J/kJ x = 6347.4 J
671.8 J + 6347.4 J = 7019.2 J
That energy will come from cooling 102.0 g H20-
7019.2 J = 102.0g x 4.18 J/g deg x (86.0 - t)
t = 69.5 deg
At this point you have 102.0 g water at 69.5 and 19 g water at 0 deg
102.0 g x 4.18 x (69.5 - t) = 19.0 x 4.18 x (t - 0)
t = final temperature = 58.6 deg C
Question 5:
A 16.7 g sample of ethanol at 230.5 K is mixed with 55.6 g of ethanol at 317.1 K. Calculate the final temperature of the mixture assuming no heat is lost to the containers and surroundings. The specific heat of ethanol is 2.44J/K·g.
Answer :
The temperature of the first solution will increase and the temperature of the second solution will decrease
16.7 g x 2.44 ( x - 230.5 ) = 55.6 g x 2.44 ( 317.1 -x)
40.75 ( x - 230.5 ) = 135.7 ( 317.1 - x )
40.75 x - 9392.9 = 43030.5 - 135.7 x
176.4 x = 52423
x = 297 K
Question 6:
Calculate the amount of heat required to completely sublime 25.0 g of solid dry ice (CO2 ) at its sublimation temperature. The heat of sublimation for carbon dioxide is 32.3 kJ/mol.
Answer :
First, determine how many moles of CO2 there are. The mol. wt. of CO2 is 12 + 2 x 16 = 44 g/mol.
So, 25.0 g equals 25/44 = 0.568 moles.
So the heat required is 0.568 mol x 32.3 kJ/mol = 18.3 kJ
Question 7:
A 188 g sample of a substance is heated to 323°C and then plunged into a 100 g aluminum calorimeter cup containing 166 g of water and a 18 g glass thermometer at 11.5°C. The final temperature is 35.3°C. What is the specific heat of the substance?
Answer :
Heat out Qiut from the substance = Q into the cup, thermometer & water
So Q out = m*c*deltaT = 0.188kg*c*(323-35.3) = 54.1c
Q in = m*c*deltaT for each substance = 0.100kg*900J/kg-oC*(35.3-11.5) + 0.166kg*4200*(35.3-11.5) + 0.018*840*(35.3-11.5) = 19165J
So 54.1c = 19165 => c = 19165/54.1 = 354J/kg-oC
Question 8:
A 448 g block of metal absorbs 5586 J of heat when its temperature changes from 19*C to 42*C.
Calculate the specific heat of the metal. Answer in units of J/kg. *C.
Answer :
delta Q = mc delta T
5586 = 0.448 x 23 x c
c = 542 J/kg.*C
Question 9:
A 500.0 g sample of an element at 195 degrees C is dropped into an ice-water mixture; 109.5 g of ice melts and an ice-water mixture remains. Calculate the specific heat of the element.
Answer :
The latent heat of fusion of ice is 334kJ/kg
Thus, Q=0.1095 kg x 334 kJ/kg = 36.573 kJ
C= 36.573 kJ / 0.5kg / 195K
C= 0.375kJ/kg.K
Question 10:
A 5.00 g sample of aluminum pellets (specific heat capacity = 0.89 J/°C·g) and a 10.00 g sample of iron pellets (specific heat capacity = 0.45 J/°C·g) are heated to 100.0°C. The mixture of hot iron and aluminum is then dropped into 98.1 g of water at 20.1°C. Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.
Answer :
Heat lost by Al.
= 5g x 0.89J/g/°C x (100°C -T) = 445 - 4.45T.
Heat lost by Fe.
= 10g x 0.45J/g/°C x (100°C - T) = 450 - 4.5T.
Heat gained by water.
= 98.1g x 4.18J/g/°C x (T - 20.1°C) = 410T - 8,242.
(445 - 4.45T) + (450 - 4.5T) = 410T - 8242
8242 + 895 = 8.95T + 410T
9137 = 419T
T = 9137/ 419 = 21.8°C Final temperature.
Tuesday, February 16, 2010
question and solution for test 1
Q2
(a) A pill consists of thicken oil layer 10 cm static on water at the height of 30 cm from the bottom part of pill. Density of oil is 600 kgm-3. (ρwater = 1000kgm-3, P atm = 1.013 x 10³ Pa). calculate:
i) Pressure gauge at the interface of oil-water (point A).
P = ρgh
= (1000)(9.8)(0.1)
= 0.98 x 103
ii) Absolute pressure at the bottom area of pill.
P = Patm + ρgh
= (1.013 x 103) + [(1000)(9.8)(0.3)]
= 3953
(b) A pressure rocket tank (figure 2) consists of 0.25 m3 kerosene weighted 200 kg. The pressure of upper part of kerosene is, Pa = 2.01 x 105
i) Density, ρ of kerosene.
P = m/v
= 200 / (0.8)(0.25)
= 1 x 104 kg/m2.
ii) Pressure at the bottom of the tank, Pb.
P = F/A
= 20k / 0.08
= 250 x 103 Nm-2.
iii) Depth of kerosene by using Pb = Pa + ρgh
Pb = Pa + ρgh
h = Pb – Pa / ρg
= 250 x 103 – 201 x 103 / (1 x 104)(9.8)
= 0.5 m.
Q3
(a) an evening news broadcast in
Day’s higher temperature was 38 ͦC. what is the corresponding value of this temperature on
i) Fahrenheit scale.
TF = (1.8)Tc + 32
= (1.8)(38) + 32
= 100.4 ͦF.
ii) Kelvin scale.
TF = Tc + 273.15
= 38 + 273.15
= 311.15 ͦK.
Temperature of the green tea is 27 ͦC. by adding an amount of ice, m ice of -25ͦ C into the water,the final temperature of the mixture is 0 ͦ C with all the ice melted and the heat capacity of the container is neglected. The latent heat of fusion for ice is Lf = 333 x 103 J.kg-1 and specific heat, c for ice and water are 2100 J.kg-1. ͦ C respectively, calculate
i) the heat change, Q of the green tea.
Q = mwater . C . ∆T
= (0.2)(4200)(27.0)
= 226803.
ii) The heat change, Q of ice in the order to warm it from temperature of
-25 ͦ C to 0 ͦ C in the form of unknown ice mass, m ice.
Q = mice . C . ∆T
mice = Q / C . ∆T
= 22680 / (2100)(0.25)
= 0.432 kg.
iii) the heat, Q that required to melt this mass of ice (in the form unknown ice mass, m ice ).
Q = mice (Lf)
= 0.432 (333 x 103)
= 143.856 x 103
= 143.86 kJ.
iv) the mass of ice , m ice when it reach thermal equilibrium. (assume there is
no heat loss).
Q = mice . C . ∆T
Mice = 143.86kJ / (2100)(0 + 25)
= 2.74 kg.
Friday, February 5, 2010
question for wave and sound
1.The processing “speed” of computer refers to the number of binary operations it can perform in one second, so it is really a frequency. If the processor of a one personal computer operates at 1.80 GHz, how much time is required for one processing cycle?
Solution
f =1.80 GHz
f = 1/T
T =1/f
=1/1.80GHZ
=1/(1.80 X 109 ) cycles /s
=5.56 X 10-10
2.humans cab hear range of frequencies from 20 to 20000Hz. To what range of wavelengths in air does this correspond (assume the sound velocity is 340 m/s)?
Solution
λ1 = v/f1 = (340 m/s) / (20/s)
= 17m
λ2 = v/f2 = (340 m/s) / (20000/s)
= 0.017m
3. A string of linear mass density 480 g/m is under a tension of 48 N. a wave of frequency 200Hz and amplitude 4.0 mm travels down the string. At what rate does the wave transport energy?
Solution
ω = 2π f = 2π(200) = 400π s-1 V = √(T/µ ) = √((48 ) N)/(0.48 kg/m = 10 m/s
P = ( ½)µvω2A2 = (0.5)(0.480 kg/m)(10 m/s)(400 πs)2(4 X 10-3m)2= 61 W
4. The G string of the mandolin is 0.34 m long and has a linear mass density of 0.004 kg/m. the thumbscrew attached to the string is adjusted to provide a tension of 71.1 N. what then is the fundamental frequency of the string?
Solution
f1= v/2L = 1/2L√(T/µ) = 1/(2)(0.34m)√((71.1 N)/ ()0.004 kg/m)
=196 Hz
5. a string wave is describe by y = 0.002 sin(0.5x – 628t). determind the amplitude, frequency,period,wavelength,and velocity of the wave.
Solution
A = 0.02m 2π/λ = 0.5 λ =12.6 m
2π/T = 628 T =0.01s f = 1 /T =100 Hz v =f λ =1260 m/s
Solution
f =1.80 GHz
f = 1/T
T =1/f
=1/1.80GHZ
=1/(1.80 X 109 ) cycles /s
=5.56 X 10-10
2.humans cab hear range of frequencies from 20 to 20000Hz. To what range of wavelengths in air does this correspond (assume the sound velocity is 340 m/s)?
Solution
λ1 = v/f1 = (340 m/s) / (20/s)
= 17m
λ2 = v/f2 = (340 m/s) / (20000/s)
= 0.017m
3. A string of linear mass density 480 g/m is under a tension of 48 N. a wave of frequency 200Hz and amplitude 4.0 mm travels down the string. At what rate does the wave transport energy?
Solution
ω = 2π f = 2π(200) = 400π s-1 V = √(T/µ ) = √((48 ) N)/(0.48 kg/m = 10 m/s
P = ( ½)µvω2A2 = (0.5)(0.480 kg/m)(10 m/s)(400 πs)2(4 X 10-3m)2= 61 W
4. The G string of the mandolin is 0.34 m long and has a linear mass density of 0.004 kg/m. the thumbscrew attached to the string is adjusted to provide a tension of 71.1 N. what then is the fundamental frequency of the string?
Solution
f1= v/2L = 1/2L√(T/µ) = 1/(2)(0.34m)√((71.1 N)/ ()0.004 kg/m)
=196 Hz
5. a string wave is describe by y = 0.002 sin(0.5x – 628t). determind the amplitude, frequency,period,wavelength,and velocity of the wave.
Solution
A = 0.02m 2π/λ = 0.5 λ =12.6 m
2π/T = 628 T =0.01s f = 1 /T =100 Hz v =f λ =1260 m/s
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