1. Water is used simply because it is cheap, commonly found and less messy. In my opinion you can as well use some cooking oil which would actually have a lower specific heat capacity, that is lower "heat inertia". It would be easier to raise the temperature of oil and thereby find a greater temperature change than water.
Your accuracy would improve on account of a greater CHANGE in temperature. You would hopefully be using a sensitive digital themometer and following the usual procedures to reduce heat losses to a minimum.
In any case the inherent errors in all heat experiments are pretty high.
2. It is very simple if you know what are the ice and steam points in the two scales of temperature.
The ice point in the Celsius scale is '0' whereas in Fahrenheit scale it is '32'
The steam point in the Celsius scale is '100' whereas in Fahrenheit scale it is '212'
Now the difference between the two points in the Celsius scale is = 100 - 0 = 100 oC
whereas in the Fahrenheit scale
it is = 212 - 32 = 180 oF
We can now write a simple equation for the conversion as follows. This is the only equation you really need to know. All other conversions and faulty thermometer problems you can work out using this equation.
(C - 0)/100 = (F - 32)/180
simplifying to
C/5 = (F - 32)/9
In your question you want the relation to convert Fahrenheit to Celsius. So making C the subject of the equation, you'll have
C = 5(F - 32)/9
For example you want find your normal body temperature. It is 98.4 oF
To convert it to Celsius from the equation above
C = 5(98.4 - 32)/9
= 36.9 0C
Get it? It's that simple. So don't try and memorize the equation. It can be confusing at times. Physics is easy if you try to learn it using the first principles like you're trying to do above.
3. This is one problem in which you’ll need a GOOD DIAGRAM to help you think it through.
Could you draw that for me please. Label the arc on the top of the cone as AB (because copper expands more than iron) and the immediate layer just below it as CD (because iron expands less than copper). Extend the cone to meet at some point which you can call ‘O’. Let the radius of this cone OC be R and the angle of the arc be angle COD = theta, the angle of deflection. This is what you want to find out right?
You can’t do this unless you know the radius of the cone OC. So let first determine R in terms of what you know in the problem, alpha’s, deltaT and diameter d.
Once you know R, it is easy to find the angle of deflection because from the same cone (imagine it to be an arc of a circle) is given by
CD = (l)[1 + (alphaFe)deltatheta] = R (theta)
Making theta that subject of the equation,
Theta = (l)[1 + (alphaFe)deltatheta]/R
To find R
You must know how much the strips expand sideways. If you add the two expansions, you’ll find
‘d’ after heating = 2d + d(deltaT)[alpahFe + alphaCu]
Neglecting the second term because it is very small, you’ll have
‘d' = 2d
The arc AB will be (l)[1 + (alphaCu)deltaT]
and arc CD will be (l)[1 + (alphaFe)deltaT]
Dividing the two equations above you’ll have
AB/CD = 1 + (alphaCu)deltaT/[1 + (alphaFe)deltaT]
You can make R the subject of the above equation and solve it. Then you substitute this in the equation for ‘theta’ and you’ve got what you were looking for.
4. The concepts involved are the priciple of the method of mixtures and mass = density x volume.
In kgl-1 density of water = 0.998 kgl-1
This means 1 l of lemonade will have a mass of 0.998 kg.
Using the principle that
heat gained by the cold object = heat lost by the hot object, you'll have
M [(2200)(10) + 333000 + (4190)(10)] = (0.998)(4190)(10)
where M is the mass of the ice = (0.917)(Vi)
because ice at - 10 oC is increased to 0 oC then from 0 oC to 0 oC and finally from 0 oC to 10 oC
Solving for M you'll get M = 0.105 kg giving you 0.115 l of ice.
Using the fact that mass cannot change
diVi = dwVw
(917)(0.115) = (998)Vw
giving you the volume of water Vw = 0.106 l
This must've been the answer you were looking for.
= [R + 2d]/R (using the values for AB and CD in terms of arc length)
and density of ice = 0.917 kgl-1