Question on themal conduction and a quick question on spectrometer

[leena19]

This problem has been SOLVED.thx.Question 1

Homework Statement

1)The temperature of an ice layer,of thickness 50m, is -50 deg.C.If the ice layer continues to grow,find the timetaken for it to grow by another 1mm.
Temperature at the bottom of ice layer is 0 deg.C

2)Calculate the minimum rate at which heat should be provided per unit area of the ice layer by warm water currents beneath to stop growth of the above mentioned layer at 50m

3)If the warm currents provide heat at the rate of 0.5Wm^-2 for 2 days,what would be the thickness of the 50m icelayer after 2 days

Homework Equations

Q/t = kA (t1-t2)/d
t=temperature
d=thickness
k=thermal conductivity of ice = 2W/m²Q/t=mL/t

L=latent heat of fusion of ice at 0 deg.C = 3.6*10⁵

The Attempt at a Solution

1) mL/t = kA (t1-t2)/d
A*1*10^-3*900*3.6*10⁵/t = 2*A*50/50 ?
I think the above equation is correct,but if the question says the thickness of the icelayer has grown by,say 40cm,then d would have to be 40 ?
or is it (40+50)/2 ? I can't quite remember which our teacher said.

2) Q/tA = k (t1-t2)/d
=2*50/50

3)I don't know

Question 2 - spectrometer

a )At the minimum deviation position,the scale reading of the spectrometer is 3degrees 16'.
If the scale reading of the spectrometer when the collimater and the telescope are kept in the same straight line ,is 223deg 46'.
Calculate the angle of minimum deviation ?

I'm not sure if I should subtract 3degrees 16' from 223deg 46' directly,or first subtract 223deg 46' from 360,then add 3degrees 16' ?b) The images of the slit due to the colours yellow,blue,red and green light are different.
Draw the relative positions of the images due to the 4 colours and also due to white light .

I've never done the spectrometer experiment in class ,so I'm not sure how the images would appear.
I know that the images of the slit for yellow and white light would appear symmetrical on the crosswires but for red,blue and green ?

 

[jbsteele]

Answer1) mL/t = kA (t1-t2)/dA*1*10-3*900*3.6*105/t = 2*A*50/50Solve for t:t = A*1*10-3*900*3.6*105/(2*A*50/50)t = 4500s = 75min2) Q/tA = k (t1-t2)/dQ/tA = 2*50/50Q/tA = 2W/m23)Q/tA = k (t1-t2)/dQ/tA = 2*50/50Q = 0.5W/m2 * 2days * 2*50/50Q = 20J/m2The thickness of the ice layer after 2 days will be the same,50m.Question 2a) Subtract 3degrees 16' from 223deg 46'.Angle of minimum deviation = 220deg 30'b) The images of the slit due to yellow and white light will appear symmetrical on the crosswires, while the red, blue and green light will appear at different positions, depending on the wavelength of each light.

 

[tomcue]

Question 1:
To solve for the time taken for the ice layer to grow by another 1mm, you can use the equation mL/t = kA (t1-t2)/d where mL is the latent heat of fusion of ice at 0 deg.C, t is the time, k is the thermal conductivity of ice, A is the cross-sectional area of the ice layer, t1 is the temperature at the bottom of the ice layer (0 deg.C), t2 is the temperature at the top of the ice layer (-50 deg.C), and d is the thickness of the ice layer (50m + 0.001m = 50.001m). So the equation would be (3.6*10^5)/(t*900) = (2*50)/(50.001). Solving for t, the time taken for the ice layer to grow by another 1mm is approximately 2000 seconds or 33.33 minutes.

To calculate the minimum rate at which heat should be provided per unit area of the ice layer to stop the growth, you can use the equation Q/tA = k (t1-t2)/d where Q is the heat provided, t is the time, A is the cross-sectional area of the ice layer, k is the thermal conductivity of ice, t1 is the temperature at the bottom of the ice layer (0 deg.C), t2 is the temperature at the top of the ice layer (-50 deg.C), and d is the thickness of the ice layer (50m). So the equation would be Q/(t*900) = 2*50/50. Solving for Q, the minimum rate at which heat should be provided per unit area is approximately 200 W/m^2.

To find the thickness of the ice layer after 2 days (172800 seconds), you can use the equation d = (Q*t*A*(t1-t2))/(k*m*L) where Q is the heat provided (0.5 W/m^2), t is the time (172800 seconds), A is the cross-sectional area of the ice layer, t1 is the temperature at the bottom of the ice layer (0 deg.C), t2 is the temperature at the top of the ice layer (-50 deg.C), k is the thermal conductivity of ice, m is the mass (density * volume) of the ice layer, and L is the latent heat of