# Temperature and Heat

1. The drawing shows a hydraulic system used with disc brakes. The force F is applied perpendicularly to the brake pedal. The brake pedal rotates about the axis shown in the drawing and causes a force to be applied perpendicularly to the input piston (radius = $9.50*10^{-3}m$) in the master cylinder. The resulting pressure is transmitted by the brake fluid to the output plungers (radii = $1.90*10^{-2}m$), which are covered with the brake linings. The linings are pressed against both sides of a disc attached to the rotating wheel. Suppose that the magnitude of F is 9.00 N. Assume that the input piston and the output plungers are at the same vertical level and find the force applied to each side of the revolving disc.

2. The brass bar and the aluminum bar in the drawing are each attached to an immovable wall. At 28°C the air gap between the rods is $1.3 10^{-3} m$. At what temperature will the gap be closed?

3. A thin spherical shell of silver has an inner radius of $2.0 x 10^{-2}m$ when the temperature is 18 degrees Celsius. The shell is heated to 147 degrees Celsius. Find the changes in the interior volume of the shell.

Answers

1. Let $F$ the force applied by the foot to the pedal and $F1$ the force applied by the pedal to the master cylinder. The momentum of the two forces with respect to the axis is equal.

$F *0.1 =F1*0.05$

$F1 =2*F$

let $F2$ the force applied to one side of the revolving disk, S1 the surface area of the master cylinder and S2 the surface area of one of the 2 lateral plungers (cylinders). The pressure applied to the master cylinder is equal to the total pressure of the plungers.

$P1 =2*P2$

$F1/S1 =2*F2/S2$

$F2 =F1 *S2/(2*S1) =2*F*R2^2/(2*R1^2) =2*9 *1.9^2*10^{-4}/(2*9.5^2*10^{-6}) = 36 N$

2. The linear dilatation of a rod obeys the law

$\Delta(L) = L*\alpha*\Delta(T)$

where $\alpha_{brass} =19*10^{-6} (Celsius)^{-1}$, $\alpha_{Aluminum} =23*10^{-6} (Celsius)^{-1}$

$\Delta(L_{brass}) +\Delta(L_{Al}) =1.3*10^{-3} m$

$L_{brass}*\alpha_{brass}*\Delta(T) +L_{Al}*\alpha_{Al}*\Delta(T) =1.3*10^{-3}$

$(2*19*10^{-6} +1*23*10^{-6})*\Delta(T) =1.3*10^-{3}$

$61*10^{-6}*delta(T) =1.3*10^{-3}$

$\Delta(T) = 21.311$ degree Celsius

$T = 28+delta(T) =28+21.3 =49.3$ degree Celsius

3. The variation of the surface S of the shell is obeying the law

$\delta(S) =S*\alpha_{Surface}*\Delta(T)$

where

$\alpha_{silverSurface} = 2*\alpha_{silverLiniar} =2*18*10^{-6} (Celsius)^{-1} =36*10^{-6} (Celsius)^{-1}$

Initial $S0 = 4*pi*R0^2 =4*pi*2*2*10^{-4} =50.26*10^{-4} m^2$

$\Delta(T) =147-18=129$ deg Celsius

$\Delta(S) =50.26*10^{-4}*36*10^{-6}*129 =2.33*10^{-5} m^2$

Final $S =S0 +\Delta(S) =(50.26+0.233)*10^{-4} =50.49*10^{-4} m^2$

The final radius of the shell is

$R =\sqrt{S/(4\pi)} =\sqrt{50.49*10^{-4}/(4\pi)} =2.0044*10^{-2} m$

The change in the interior volume is $V-V0 =(4*\pi)/3 *(R^3-R0^3) =(4*\pi)/3 *(2.0044^3*10^{-6} -2^3*10^{-6}) =2.2*10^{-7} m^3$