An aluminum can is ﬁlled to the brim with a liquid. The can and the liquid are heated so their temperatures change by the same amount. The can’s initial volume at 5 oC is 3.5×10−4 m3. The coefﬁcient of volume expansion for aluminum is 69×10−6(Co)−1. When the can and the liquid are heated to 78 oC, 3.6×10−6m3 of liquid spills over. What is the coefﬁcient of volume expansion of the liquid?

Question

An aluminum can is ﬁlled to the brim with a liquid. The can and the liquid are heated so their temperatures change by the same amount. The can’s initial volume at 5 oC is 3.5×10−4 m3. The coefﬁcient of volume expansion for aluminum is 69×10−6(Co)−1. When the can and the liquid are heated to 78 oC, 3.6×10−6m3 of liquid spills over. What is the coefﬁcient of volume expansion of the liquid?

Answer 1

Initial volume of liquid, V

= 3.5 × 10-4 m³

Change in temperature, ΔT
=(78-5)=73°

Spillage, ΔV
=3.6 × 10-6 m³

Coefficient of volume expansion over that of the can
=(ΔV/V)/ΔT
=(3.6*10^(-6)/(3.5*10^(-4)))/(78-5)
=1.4090*10^(-4)

Coefficient of volume expansion of can, Cc
= 69*10^(-6)

Coefficient of volume expansion of liquid
=1.4090*10^(-4)+Cc
=2.099*10^(-4)
=2.1*10^(-4) approx.

Answer 2

To determine the coefficient of volume expansion of the liquid, we need to use the principle of conservation of volume.

The initial volume of the can (V_initial) is given as 3.5 × 10^(-4) m^3, and the temperature change (ΔT) is the same for both the can and the liquid.

We can use the formula for the volume change of a substance due to temperature change:

ΔV = V_initial * β * ΔT

where ΔV is the change in volume, V_initial is the initial volume, β is the coefficient of volume expansion, and ΔT is the temperature change.

In this case, we can calculate the volume change of the can using the given information:

ΔV_can = V_initial * β_can * ΔT

Since the temperature change is the same for both the can and the liquid, we can also calculate the volume change of the liquid:

ΔV_liquid = V_initial * β_liquid * ΔT

Now, to determine the coefficient of volume expansion of the liquid, we need to isolate β_liquid.

From the given information, we know that when heated to 78 oC, the liquid spills over by a volume of 3.6 × 10^(-6) m^3. So, we have:

ΔV_liquid = 3.6 × 10^(-6) m^3

Substituting the known values into the formula:

3.6 × 10^(-6) m^3 = V_initial * β_liquid * ΔT

Because the volume change of the can and the liquid comes from the same initial volume, we can write:

ΔV = ΔV_can + ΔV_liquid

Since the volume change of the can is not given, we can express it in terms of the spilled liquid volume:

ΔV_can = -ΔV_liquid

Now, let's substitute this into our earlier equation:

ΔV_liquid = -ΔV_liquid + ΔV_liquid

Simplifying:

2ΔV_liquid = ΔV_liquid

ΔV_liquid = (1/2) * ΔV_liquid

Considering the given information:

3.6 × 10^(-6) m^3 = (1/2) * ΔV_liquid

Multiplying both sides by 2:

7.2 × 10^(-6) m^3 = ΔV_liquid

Finally, substituting this value back into the formula for volume change of the liquid:

7.2 × 10^(-6) m^3 = V_initial * β_liquid * ΔT

Rearranging to solve for β_liquid:

β_liquid = (7.2 × 10^(-6) m^3) / (V_initial * ΔT)

Now you can plug in the given values for V_initial (3.5 × 10^(-4) m^3) and ΔT (78 oC - 5 oC) to calculate β_liquid.