Question

A steel container tube (open at the top) is filled to the brim with 500.00 mL (0.790 kg) of ethyl alcohol at 20.0 oC. A medical student warms the tube in her hands. The final equilibrium temperature of the steel and alcohol is 34.0 oC. Physical data for steel and alcohol appear in the attached data sheet. Which of the following is most likely to occur?

1. (a) The level of alcohol in the tube will appear to stay the same.

2. (b) Approximately 0.52 mL of the alcohol will spill out of the tube.

3. (c) The level of alcohol in the tube will drop below the rim of the tube by 0.52 mL.

4. (d) Approximately 0.77 mL of the alcohol will spill out of the tube.

5. (e) The level of alcohol in the tube will drop below the rim of the tube by 0.77 mL.

Answer 1

The coefficient of volume expansion β for steel is 35 × 10^{– 6}C^-1 and alcohol is 110 × 10^{– 6}  C^-1 (Standard values)

The change in volume ΔV is given bt ΔV = βVΔT.

The steel and alcohol increase in volume, but the alcohol increases more(as the coefficient of volume expansion of alcohol is more), so the alcohol will spill out and the amount spilled is the difference in their volume changes. We can use the equation for volume expansion to calculate the change in volume of the steel tube and alcohol.

V_spill=ΔValcohol − ΔV_steel.

V_spill=(β_alcohol−β_steel)VΔT.

V = 500 mL, ΔT= 34-20 = 14 C.

So, V_spill = (110 × 10^{– 6} -  35 × 10^{– 6} ) x 500 x 14 = .77 ml

Hence option (d)