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Describe the isothermal process. A right rectangular prism of cast iron has dimensions of 55 × 45 × 35 mm at 25 °C. The prism is heated to 88 °C. The coefficient of linear expansion of cast iron is 9 × 10⁻⁶/°C. Calculate the following: 7.2.1 The increase in volume 7.2.2 The volume at 88 °C

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Here are the solutions to Question 7:

QUESTION 7: HEAT

7.1 Describe the isothermal process.

An isothermal process is a thermodynamic process during which the temperature of a system remains constant. This occurs when the system is in thermal contact with a large heat reservoir, allowing heat to be exchanged to maintain a steady temperature.

7.2 A right rectangular prism of cast iron has dimensions of 55×45×35 mm55 \times 45 \times 35 \text{ mm} at 25∘C25^\circ \text{C}. The prism is heated to 88∘C88^\circ \text{C}. The coefficient of linear expansion of cast iron is 9×10−6/∘C9 \times 10^{-6}/^\circ \text{C}.

Given:

  • Initial dimensions: L0=55 mmL_0 = 55 \text{ mm}, W0=45 mmW_0 = 45 \text{ mm}, H0=35 mmH_0 = 35 \text{ mm}
  • Initial temperature, T1=25∘CT_1 = 25^\circ \text{C}
  • Final temperature, T2=88∘CT_2 = 88^\circ \text{C}
  • Coefficient of linear expansion, α=9×10−6/∘C\alpha = 9 \times 10^{-6}/^\circ \text{C}

7.2.1 The increase in volume

  1. Calculate the original volume (V0V_0).

    V0=L0×W0×H0V_0 = L_0 \times W_0 \times H_0 V0=55 mm×45 mm×35 mmV_0 = 55 \text{ mm} \times 45 \text{ mm} \times 35 \text{ mm} V0=86625 mm3V_0 = 86625 \text{ mm}^3
  2. Calculate the change in temperature (ΔT\Delta T).

    ΔT=T2−T1\Delta T = T_2 - T_1 ΔT=88∘C−25∘C=63∘C\Delta T = 88^\circ \text{C} - 25^\circ \text{C} = 63^\circ \text{C}
  3. Calculate the coefficient of volumetric expansion (γ\gamma).

    For isotropic materials, the coefficient of volumetric expansion is approximately three times the coefficient of linear expansion.

    γ=3α\gamma = 3\alpha γ=3×(9×10−6/∘C)=27×10−6/∘C\gamma = 3 \times (9 \times 10^{-6}/^\circ \text{C}) = 27 \times 10^{-6}/^\circ \text{C}
  4. Calculate the increase in volume (ΔV\Delta V).

    ΔV=V0γΔT\Delta V = V_0 \gamma \Delta T ΔV=(86625 mm3)×(27×10−6/∘C)×(63∘C)\Delta V = (86625 \text{ mm}^3) \times (27 \times 10^{-6}/^\circ \text{C}) \times (63^\circ \text{C}) ΔV=147.303375 mm3\Delta V = 147.303375 \text{ mm}^3

    The increase in volume is 147.30 mm3\boxed{\text{147.30 mm}^3}.

    7.2.2 The volume at 88∘C88^\circ \text{C}

  5. Calculate the final volume (VfV_f).

    Vf=V0+ΔVV_f = V_0 + \Delta V Vf=86625 mm3+147.303375 mm3V_f = 86625 \text{ mm}^3 + 147.303375 \text{ mm}^3 Vf=86772.303375 mm3V_f = 86772.303375 \text{ mm}^3

    The volume at 88∘C88^\circ \text{C} is 86772.30 mm3\boxed{\text{86772.30 mm}^3}.

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