A ‘thermal tap’ used in a certain apparatus consists of a silica rod which fits tightly inside an aluminium tube whose internal diameter is 8mm at

0

0C. When the temperature is raised, the fit is no longer exact. Calculate

what change in temperature is necessary to produce a channel whose

cross-section is equal to that of a tube of 1mm internal diameter. (linear

expansivity of silica = 8X10-6

/K and linear expansivity of aluminium =

26X10-6

/K )

Correct question is;

A thermal tap used in a certain apparatus consists of a silica rod which fits tightly inside an aluminium tube whose internal diameter is 8mm at 0°C.When the temperature is raised ,the fits is no longer exact. Calculate what change in temperature is necessary to produce a channel whose cross-sectional is equal to that of the tube of 1mm. (linear expansivity of silica = 8 × 10^(-6) /K and linear expansivity of aluminium = 26 × 10^(-6) /K).

ΔT = 268.67K

Explanation:

We are given;

d1 = 8mm

d2 = 1mm

At standard temperature and pressure conditions, the temperature is 273K.

Thus; Initial temperature; T1 = 273K,

Using the combined gas law, we have;

P1×V1/T1 = P2×V2/T2

The pressure is constant and so P1 = P2. They will cancel out in the combined gas law to give:

V1/T1 = V2/T2

Now, volume of the tube is given by the formula;V = Area × height = Ah

Thus;

V1 = (πd1²/4)h

V2 = (π(d2)²/4)h

Thus;

(πd1²/4)h/T1 = (π(d2)²/4)h/T2

π, h and 4 will cancel out to give;

d1²/T1 = (d2)²/T2

T2 = ((d2)² × T1)/d1²

T2 = (1² × T1)/8²

T2 = 273/64

T2 = 4.23K

Therefore, Change in temperature is; ΔT = T2 - T1

ΔT = 273 - 4.23

ΔT = 268.67K

Thus, the temperature decreased to 268.67K

the answer is 12 as i may remember

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