物理高手 大一普物解題 - 三鶯
By Charlie
at 2011-11-29T01:59
at 2011-11-29T01:59
Table of Contents
路過手癢,答案作法不保證正確
請自行查證~謝謝!歡迎板友不吝指教!
就到第 7題,我欠一堆讀書債沒還
剩下請其他板友協助
1.http://ppt.cc/zkJj
0→75 Celsius degree
L'=(1+0.0012)L
∵(1+ε)^3≒1+3ε
∴V'=(1+0.0012*3)V
D =M/V
→D'=M/V'=M/(1+0.0012*3)V≒(1-0.0012*3)D →0.36%
2.http://ppt.cc/Ddfw
(a)0→40min
△H=0.569*3031*(290-280)=K*40
∴Absorption rate(K)=0.569*3031*0.25 J/min
Fusion energy(K_f)
0.569*K_f=K*30,K_f=22732.5 J/kg
(b)70→90min
Specific heat(frozen phase)=S_f
△H=K*20=0.569*(280-260)*S_f
S_f=757.75 J/kg.K
3.http://ppt.cc/fGmH
Vaporation heat of water=539 Cal/g
Environment transfer is negletabl; thus have the following:
△H=380*(T-100)*0.0923
=120*(100-20)*0.0923+ 120*(100-20)*1+ 539*7.26
copper bowl water
For water :120*(100-20)*1+ 539*7.26=13513.14
For copper bowl:120*(100-20)*0.0923 =886.08
T=511 Celsius degree
4.http://ppt.cc/MPFR
Application of law of thermodynamics
△E=△Q- △W=△Q'- △W'
(a)Path iaf=Path ibf →50.5- 21.4=37.0- W'
W'=7.9 Cal
(b)Path fi=-(Path iaf)→For the opposite direction
∴-(50.5- 21.4)=14.2+ Q'
Q'=-43.3 Cal
(c)E_int(f)- E_int(i)=Q- W →First law!
∴E_int(f)=E_int(i)+ Q- W=40.5 Cal
(d)E_int(b)=22.5
@For path ib, it an iso-pressure procedure
All energy is used as W(△E=0), Q_ib=W_ib=11.1
@For path bf, it an iso-volume procedure
All energy is turned to be internal energy, Q_bf=E_bf=18.0
5.http://ppt.cc/CcUO
nRT
For isothermal process △E=0, P=-----
V
╭ 0.31 ╭0.31 nRT 0.31
∴W=△Q=│ PdV=│ -----dV=nRT.ln(----------)
╯ 0.31*2/3 ╯0.31*2/3 V 0.31*2/3
6
1280*---=0.9*8.314*0.4054*T →T=351.6 K
5
6.http://ppt.cc/6~yO
Application of law of thermodynamics
△E=△Q- △W=△Q'- △W'
(a)23.3=△E+ P△V=△E+ (1.1*1.01*10^5)*(25*10^-6)
For 1 atm=1.01*10^5 Nt/m^2
→△E=23.3- 2.7775=20.5 J
(b)n=3.8*10^-3 mol
@Constant P, △Q=△E+ P△V=△E+ nR△T
2.7775= (3.8*10^-3)*8.314*△T, △T=87.9
23.3
∴C_p=-------------------=69.8 J/mol.K
87.9*(3.8*10^-3)
@Constant V, △Q=△E
∴C_v=69.8- 8.314=61.5 J/mol.K
7.http://ppt.cc/LuIf
dW= dQ- dE
3
→dW= TdS-(---nR)dT
2
7 2
According to the figure: T=---T_s- ---S
6 3
7 2 3
→dW=(---T_s- ---S)dS-(---nR)dT
6 3 2
╭13.9 7 2 ╭449 3
→W=│ (---T_s- ---S)dS-│ (---nR)dT
╯1/4*13.9 6 3 ╯1/2*449 2
→W=-1095 J
--
請自行查證~謝謝!歡迎板友不吝指教!
就到第 7題,我欠一堆讀書債沒還
剩下請其他板友協助
1.http://ppt.cc/zkJj
0→75 Celsius degree
L'=(1+0.0012)L
∵(1+ε)^3≒1+3ε
∴V'=(1+0.0012*3)V
D =M/V
→D'=M/V'=M/(1+0.0012*3)V≒(1-0.0012*3)D →0.36%
2.http://ppt.cc/Ddfw
(a)0→40min
△H=0.569*3031*(290-280)=K*40
∴Absorption rate(K)=0.569*3031*0.25 J/min
Fusion energy(K_f)
0.569*K_f=K*30,K_f=22732.5 J/kg
(b)70→90min
Specific heat(frozen phase)=S_f
△H=K*20=0.569*(280-260)*S_f
S_f=757.75 J/kg.K
3.http://ppt.cc/fGmH
Vaporation heat of water=539 Cal/g
Environment transfer is negletabl; thus have the following:
△H=380*(T-100)*0.0923
=120*(100-20)*0.0923+ 120*(100-20)*1+ 539*7.26
copper bowl water
For water :120*(100-20)*1+ 539*7.26=13513.14
For copper bowl:120*(100-20)*0.0923 =886.08
T=511 Celsius degree
4.http://ppt.cc/MPFR
Application of law of thermodynamics
△E=△Q- △W=△Q'- △W'
(a)Path iaf=Path ibf →50.5- 21.4=37.0- W'
W'=7.9 Cal
(b)Path fi=-(Path iaf)→For the opposite direction
∴-(50.5- 21.4)=14.2+ Q'
Q'=-43.3 Cal
(c)E_int(f)- E_int(i)=Q- W →First law!
∴E_int(f)=E_int(i)+ Q- W=40.5 Cal
(d)E_int(b)=22.5
@For path ib, it an iso-pressure procedure
All energy is used as W(△E=0), Q_ib=W_ib=11.1
@For path bf, it an iso-volume procedure
All energy is turned to be internal energy, Q_bf=E_bf=18.0
5.http://ppt.cc/CcUO
nRT
For isothermal process △E=0, P=-----
V
╭ 0.31 ╭0.31 nRT 0.31
∴W=△Q=│ PdV=│ -----dV=nRT.ln(----------)
╯ 0.31*2/3 ╯0.31*2/3 V 0.31*2/3
6
1280*---=0.9*8.314*0.4054*T →T=351.6 K
5
6.http://ppt.cc/6~yO
Application of law of thermodynamics
△E=△Q- △W=△Q'- △W'
(a)23.3=△E+ P△V=△E+ (1.1*1.01*10^5)*(25*10^-6)
For 1 atm=1.01*10^5 Nt/m^2
→△E=23.3- 2.7775=20.5 J
(b)n=3.8*10^-3 mol
@Constant P, △Q=△E+ P△V=△E+ nR△T
2.7775= (3.8*10^-3)*8.314*△T, △T=87.9
23.3
∴C_p=-------------------=69.8 J/mol.K
87.9*(3.8*10^-3)
@Constant V, △Q=△E
∴C_v=69.8- 8.314=61.5 J/mol.K
7.http://ppt.cc/LuIf
dW= dQ- dE
3
→dW= TdS-(---nR)dT
2
7 2
According to the figure: T=---T_s- ---S
6 3
7 2 3
→dW=(---T_s- ---S)dS-(---nR)dT
6 3 2
╭13.9 7 2 ╭449 3
→W=│ (---T_s- ---S)dS-│ (---nR)dT
╯1/4*13.9 6 3 ╯1/2*449 2
→W=-1095 J
--
Tags:
三鶯
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