Answer :
Final answer:
The volume of turpentine that overflows when the system is heated to 80°C is 105.1 cm^3. This is calculated by determining the thermal expansion of both the aluminium cylinder and the turpentine, and subtracting the former from the latter.
Explanation:
This question is essentially about the thermal expansion of aluminum and turpentine. We can first calculate the expansion of the aluminum cylinder and then the expansion of the turpentine, since the turpentine expands more than the cylinder, it will overflow.
The change in volume ∆V of the aluminium cylinder due to heating can be approximated using the formula: ∆V = V_initial * β * ∆T , where V_initial is the initial volume, β is the volume expansion coefficient, and ∆T is the change in temperature. For the cylinder, V_initial = 2000 cm^3 (as 2L=2000cm^3), β = 24×10^−6 C^−1 and ∆T = 80−20 = 60∘C. Plugging the numbers we find that the volume of the cylinder increases by about 2.88 cm^3. Similarly, for the turpentine, V_initial = 2000 cm^3, β = 9×10^−4 C^−1 and ∆T = 60∘C. The turpentine therefore expands by about 108 cm^3.
However, since the cylinder also expands, the net overflow of turpentine will be the expansion of the turpentine minus the expansion of the cylinder, or 108 cm^3 - 2.88 cm^3 = 105.1 cm^3. So the correct choice is b) 105.1 cm^3.
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