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the cavity within a copper β = 51 × 10⁻⁶ (c⁰)⁻¹ sphere has a volume of 1.140 × 10⁻³ m³. into this cavity is placed 1.100 × 10⁻³ m³ of benzene β = 1240 × 10⁻⁶ (c⁰)⁻¹. both the copper and the benzene have the same temperature. by what amount δt should the temperature of the sphere and the benzene within it be increased, so that the liquid just begins to spill out?
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Explanation:

Step1: Write the volume expansion formula

The volume expansion formula is \( \Delta V = V_0\beta\Delta T\). For the benzene, \( \Delta V_{benzene}=V_{0,benzene}\beta_{benzene}\Delta T\). For the copper cavity, \( \Delta V_{copper}=V_{0,copper}\beta_{copper}\Delta T\). When the liquid just begins to spill out, \( \Delta V_{benzene}-\Delta V_{copper}=V_{0,copper} - V_{0,benzene}\)

Step2: Substitute the formulas into the spill - out condition

\(V_{0,benzene}\beta_{benzene}\Delta T-V_{0,copper}\beta_{copper}\Delta T=V_{0,copper} - V_{0,benzene}\)

\(\Delta T=\frac{V_{0,copper} - V_{0,benzene}}{V_{0,benzene}\beta_{benzene}-V_{0,copper}\beta_{copper}}\)

Given \(V_{0,copper}=1.140\times 10^{-3}\space m^{3}\), \(V_{0,benzene}=1.100\times 10^{-3}\space m^{3}\), \(\beta_{copper}=51\times 10^{-6}(C^{\circ})^{-1}\), \(\beta_{benzene}=1240\times 10^{-6}(C^{\circ})^{-1}\)

Substitute the values:

$$
LATEXBLOCK0
$$

Answer:

\(\Delta T = 30.6^{\circ}C\)