Question

The cavity within a copper [β = 51 × 10-6 (C°)-1] sphere has a volume of 1.150 × 10-3 m3. Into this cavity is placed 1.100 × 10-3 m3 of benzene [β = 1240 × 10-6 (C°)-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?

Answer 1

Answer on Question #37953, Physics, Other

Question:

The cavity within a copper [β = 51 × 10⁻⁶ (C°)-1] sphere has a volume of 1.150 × 10⁻³ m³. Into this cavity is placed 1.100 × 10⁻³ m³ of benzene [β = 1240 × 10⁻⁶ (C°)-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?

Answer:

The volume of body when temperature change can be expressed as:

$$V = V_0 (1 + \beta \Delta T)$$

where $V_0$ is initial volume, $\beta$ is volumetric temperature expansion coefficient, $\Delta T$ is change of temperature.

The liquid just begins to spill out if volume of benzene equals volume of cavity:

$$V_c (1 + \beta_c \Delta T) = V_b (1 + \beta_b \Delta T)$$

where $V_c$ and $V_b$ are initial volumes of cavity and benzene

Therefore:

$$\Delta T = \frac{V_c - V_b}{\beta_b - \beta_c} = \frac{1.150 - 1.100}{1240 - 51} \cdot 10^3 = 0.04205{}^\circ \text{C}$$

Answer: 0.04205 °C