Q1:
A heater provides 5000J of energy in 2 minutes. Find the power of the heater
A1:
P=E/t
P=5000/2x60
P=41┘2┘3 Js^-1 (Ans. 41┘2┘3 W)
咁我想問下, 我個unit應該寫W定係Js^-1? 因為1W=1Js^-1...

Q2:
Js^-1, W, kWh, Jkg^-1 0C^-1 有咩分別? 幾時用邊d?

Q3:
1kg of water @ 90degree Celsius is added to 2kg of water @ 20degree Celsius. Find the temperature of the mixture

Q4:
0.2kg of mile @ 60degree Celsius is poured into a 0.5-kg cup at room temp. 25degree Celsius. If the final temperature of the mixture is 52degree Celsius, find the specific heat capacity of the cup. Assume the specific heat capacity of milk is 4000 Jkg^-1 0C^-1

Q5:
A light ray moves from air to glass (refractive index 1.5) with angle of incidence 35. Calculate the angle of refraction

Q6:
A light ray moves from water to air with angle of incidence 10. Calculate the angle of refraction

Q7:
我想知道幾時係用 Ng=sin(Pheta)a/sin(Pheta)g; 幾時用 N1sin(Pheta)1=N1sin(Pheta)2
因為我有時d數唔知點解會計唔岩... 但都係Snell's law者... sosad


thx

tridentenviro :Q1:
A heater provides 5000J of energy in 2 minutes. Find the power of the heater
A1:
P=E/t
P=5000/2x60
P=41┘2┘3 Js^-1 (Ans. 41┘2┘3 W)
咁我想問下, 我個unit應該寫W定係Js^-1? 因為1W=1Js^-1...

Q2:
Js^-1, W, kWh, Jkg^-1 0C^-1 有咩分別? 幾時用邊d?

Q3:
1kg of water @ 90degree Celsius is added to 2kg of water @ 20degree Celsius. Find the temperature of the mixture

Q4:
0.2kg of mile @ 60degree Celsius is poured into a 0.5-kg cup at room temp. 25degree Celsius. If the final temperature of the mixture is 52degree Celsius, find the specific heat capacity of the cup. Assume the specific heat capacity of milk is 4000 Jkg^-1 0C^-1


thx

Q1: 我學校教寫W
Q2: W is the unit of power
1kW = 1000W x 3600s (k=1000, 1hour have 60x60=3600 seconds)
= 3,600,000 J
l = symbol for latent heat, E = energy, m = mass
E = ml
l = E/m
Unit of latent heat= unit of E / unit of m = J/kg = Jkg^-1

E=energy, m=mass, △T=change in temperature
E=mc△T
Unit of c = unit of Q/(unit of m x unit of △T) = J/ (kg x degree Celsius) = Jkg^-10C^-1


Q3:
mc△T = mc△T
(1)(4200)(90-T) = (2)(4200)(T-20)
T= 50 degree Celsius

Q4:
mc△T = mc△T
(0.2)(4000)(60-52) = (0.5)(c)(52-25)
c = 497.7777777777
c = 498 Jkg^-1

Are they correct ??

Q5:
A light ray moves from air to glass (refractive index 1.5) with angle of incidence 35. Calculate the angle of refraction

By Snell's Law
(n1)sin(Pheta1) = (n2)sin(Pheta2)
(1)sin35'=(1.5)sin(angle of refraction)
angle of refraction = 22.481' = 28.4'

Q6:
A light ray moves from water to air with angle of incidence 10. Calculate the angle of refraction
Refractive Index of Water = 1.33
By Snell's Law
(n1)sin(Pheta1) = (n2)sin(Pheta2)
(1.33)sin10'=(1)sin(angle of refraction)
angle of refraction = 13.353' = 13.4'

Q7:
我想知道幾時係用 Ng=sin(Pheta)a/sin(Pheta)g; 幾時用 N1sin(Pheta)1=N1sin(Pheta)2
因為我有時d數唔知點解會計唔岩... 但都係Snell's law者..

兩者都係一樣 =] (我地學校個阿 sir F.3 講 ge)
雖然遵記 C.W. Sham 話係 AL 嘢,
但係我一直用(n1)sin(Pheta1) = (n2)sin(Pheta2)嚟做數 ...

tridentenviro :Q1:
A heater provides 5000J of energy in 2 minutes. Find the power of the heater
A1:
P=E/t
P=5000/2x60
P=41┘2┘3 Js^-1 (Ans. 41┘2┘3 W)
咁我想問下, 我個unit應該寫W定係Js^-1? 因為1W=1Js^-1..

W 同 Js^-1 唔係互通架咩 ?

tridentenviro :Q7:
我想知道幾時係用 Ng=sin(Pheta)a/sin(Pheta)g; 幾時用 N1sin(Pheta)1=N1sin(Pheta)2
因為我有時d數唔知點解會計唔岩... 但都係Snell's law者... sosad

thx

前者只可計算光線由空氣入物質，後者則可計算光線由物質入物質。
因此用後者公式會比前者公式方便。易記與否則見人見智。