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A glass bulb is filled with 0,6 m³ of mercury at a temperature of 20 °C. The bulb and mercury are heated to 70 °C. The linear coefficient of expansion of glass is 11 × 10⁻⁶/°C and the volume coefficient of expansion of mercury is 20 × 10⁻⁵/°C. Calculate the overflow of mercury.

Physics

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Answer

0.00501 m3\text{0.00501 m}^3

Step-by-step solution

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Here's the solution for question 3:

Given:
Initial volume (V0V_0) = 0.6 m30.6 \text{ m}^3
Initial temperature (T0T_0) = 20 °C20 \text{ °C}
Final temperature (TfT_f) = 70 °C70 \text{ °C}
Linear coefficient of expansion of glass (αg\alpha_g) = 11×10−6 /°C11 \times 10^{-6} \text{ /°C}
Volume coefficient of expansion of mercury (γm\gamma_m) = 20×10−5 /°C20 \times 10^{-5} \text{ /°C}

  1. Calculate the change in temperature (ΔT\Delta T).

    ΔT=Tf−T0=70 °C−20 °C=50 °C\Delta T = T_f - T_0 = 70 \text{ °C} - 20 \text{ °C} = 50 \text{ °C}
  2. Calculate the volumetric expansion coefficient of glass (γg\gamma_g).

    The volumetric expansion coefficient for a solid is approximately three times its linear expansion coefficient.

    γg=3αg=3×(11×10−6 /°C)=33×10−6 /°C\gamma_g = 3 \alpha_g = 3 \times (11 \times 10^{-6} \text{ /°C}) = 33 \times 10^{-6} \text{ /°C}
  3. Calculate the change in volume of the glass bulb (ΔVg\Delta V_g).

    ΔVg=V0γgΔT\Delta V_g = V_0 \gamma_g \Delta T ΔVg=(0.6 m3)×(33×10−6 /°C)×(50 °C)\Delta V_g = (0.6 \text{ m}^3) \times (33 \times 10^{-6} \text{ /°C}) \times (50 \text{ °C}) ΔVg=0.00099 m3\Delta V_g = 0.00099 \text{ m}^3
  4. Calculate the change in volume of the mercury (ΔVm\Delta V_m).

    ΔVm=V0γmΔT\Delta V_m = V_0 \gamma_m \Delta T ΔVm=(0.6 m3)×(20×10−5 /°C)×(50 °C)\Delta V_m = (0.6 \text{ m}^3) \times (20 \times 10^{-5} \text{ /°C}) \times (50 \text{ °C}) ΔVm=(0.6 m3)×(200×10−6 /°C)×(50 °C)\Delta V_m = (0.6 \text{ m}^3) \times (200 \times 10^{-6} \text{ /°C}) \times (50 \text{ °C}) ΔVm=0.006 m3\Delta V_m = 0.006 \text{ m}^3
  5. Calculate the overflow of mercury.

    The overflow is the difference between the absolute expansion of mercury and the expansion of the glass bulb.

    Overflow=ΔVm−ΔVg\text{Overflow} = \Delta V_m - \Delta V_g Overflow=0.006 m3−0.00099 m3\text{Overflow} = 0.006 \text{ m}^3 - 0.00099 \text{ m}^3 Overflow=0.00501 m3\text{Overflow} = \boxed{\text{0.00501 m}^3}
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