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An iron plate is a 10 cm square at 12 °C. The linear expansion coefficient of iron is 11 x 10 -6/°C. Calculate: a) The area expansion coefficient of iron. b) The lengths of the sides at 51 °C. c) The new area at 51 °C.

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4. An iron plate is a 10 cm square at 12 °C. The linear expansion coefficient of iron is 11×10−6 /°C11 \times 10^{-6} \text{ /°C}. Calculate:

a) The area expansion coefficient of iron.

  1. The area expansion coefficient (β\beta) is approximately twice the linear expansion coefficient (α\alpha).

    β=2α\beta = 2\alpha

  2. Substitute the given linear expansion coefficient.

    β=2×(11×10−6 /°C)\beta = 2 \times (11 \times 10^{-6} \text{ /°C}) β=22×10−6 /°C\beta = 22 \times 10^{-6} \text{ /°C}

    The area expansion coefficient of iron is \boxed{\text{22 \times 10^{-6} /°C}}.

    b) The lengths of the sides at 51 °C.

  3. Identify the initial length, initial temperature, final temperature, and linear expansion coefficient.

    Calculate the change in temperature (ΔT\Delta T).
    Initial length, L1=10 cmL_1 = 10 \text{ cm}
    Initial temperature, T1=12 °CT_1 = 12 \text{ °C}
    Final temperature, T2=51 °CT_2 = 51 \text{ °C}
    Linear expansion coefficient, α=11×10−6 /°C\alpha = 11 \times 10^{-6} \text{ /°C}

    ΔT=T2−T1=51 °C−12 °C=39 °C\Delta T = T_2 - T_1 = 51 \text{ °C} - 12 \text{ °C} = 39 \text{ °C}

  4. Use the formula for linear thermal expansion to find the final length (L2L_2).

    L2=L1(1+αΔT)L_2 = L_1 (1 + \alpha \Delta T) L2=10 cm×(1+(11×10−6 /°C)×(39 °C))L_2 = 10 \text{ cm} \times (1 + (11 \times 10^{-6} \text{ /°C}) \times (39 \text{ °C})) L2=10 cm×(1+0.000429)L_2 = 10 \text{ cm} \times (1 + 0.000429) L2=10 cm×1.000429L_2 = 10 \text{ cm} \times 1.000429 L2=10.00429 cmL_2 = 10.00429 \text{ cm}

    The lengths of the sides at 51 °C are 10.00429 cm\boxed{\text{10.00429 cm}}.

    c) The new area at 51 °C.

  5. Calculate the initial area (A1A_1).

    A1=L12=(10 cm)2=100 cm2A_1 = L_1^2 = (10 \text{ cm})^2 = 100 \text{ cm}^2

  6. Use the formula for area thermal expansion to find the new area (A2A_2).

    A2=A1(1+βΔT)A_2 = A_1 (1 + \beta \Delta T) A2=100 cm2×(1+(22×10−6 /°C)×(39 °C))A_2 = 100 \text{ cm}^2 \times (1 + (22 \times 10^{-6} \text{ /°C}) \times (39 \text{ °C})) A2=100 cm2×(1+0.000858)A_2 = 100 \text{ cm}^2 \times (1 + 0.000858) A2=100 cm2×1.000858A_2 = 100 \text{ cm}^2 \times 1.000858 A2=100.0858 cm2A_2 = 100.0858 \text{ cm}^2

    The new area at 51 °C is 100.0858 cm2\boxed{\text{100.0858 cm}^2}.

    5. A metal plate is a 1 m square at 283 K. There is a hole of 600 mm in diameter in the centre. The linear coefficient of expansion of the metal is 12.5×10−6 /K12.5 \times 10^{-6} \text{ /K}. Calculate:

    a) The temperature of the metal plate if it is heated until the sides are 1.005 m long.

  7. Identify the given values.

    Initial length, L1=1 mL_1 = 1 \text{ m}
    Initial temperature, T1=283 KT_1 = 283 \text{ K}
    Final length, L2=1.005 mL_2 = 1.005 \text{ m}
    Linear expansion coefficient, α=12.5×10−6 /K\alpha = 12.5 \times 10^{-6} \text{ /K}

  8. Use the linear thermal expansion formula to find the change in temperature (ΔT\Delta T).

    L2=L1(1+αΔT)L_2 = L_1 (1 + \alpha \Delta T) 1.005 m=1 m×(1+(12.5×10−6 /K)×ΔT)1.005 \text{ m} = 1 \text{ m} \times (1 + (12.5 \times 10^{-6} \text{ /K}) \times \Delta T) 1.005=1+(12.5×10−6)ΔT1.005 = 1 + (12.5 \times 10^{-6}) \Delta T 0.005=(12.5×10−6)ΔT0.005 = (12.5 \times 10^{-6}) \Delta T ΔT=0.00512.5×10−6=400 K\Delta T = \frac{0.005}{12.5 \times 10^{-6}} = 400 \text{ K}

  9. Calculate the final temperature (T2T_2).

    T2=T1+ΔTT_2 = T_1 + \Delta T T2=283 K+400 KT_2 = 283 \text{ K} + 400 \text{ K} T2=683 KT_2 = 683 \text{ K}

    The temperature of the metal plate is 683 K\boxed{\text{683 K}}.

    b) The diameter of the hole at this temperature.

  10. Identify the initial diameter of the hole and the change in temperature from part a).

    Initial diameter, D1=600 mm=0.6 mD_1 = 600 \text{ mm} = 0.6 \text{ m}
    ΔT=400 K\Delta T = 400 \text{ K}
    Linear expansion coefficient, α=12.5×10−6 /K\alpha = 12.5 \times 10^{-6} \text{ /K}

  11. Use the linear thermal expansion formula to find the final diameter (D2D_2).

    D2=D1(1+αΔT)D_2 = D_1 (1 + \alpha \Delta T) D2=0.6 m×(1+(12.5×10−6 /K)×(400 K))D_2 = 0.6 \text{ m} \times (1 + (12.5 \times 10^{-6} \text{ /K}) \times (400 \text{ K})) D2=0.6 m×(1+0.005)D_2 = 0.6 \text{ m} \times (1 + 0.005) D2=0.6 m×1.005D_2 = 0.6 \text{ m} \times 1.005 D2=0.603 mD_2 = 0.603 \text{ m}

    The diameter of the hole at this temperature is 0.603 m\boxed{\text{0.603 m}}.

    c) The increase in area of the hole at this temperature.

  12. Calculate the initial area of the hole (Ahole,1A_{hole,1}).

    Ahole,1=π(D12)2=π(0.6 m2)2=π(0.3 m)2=0.09π m2A_{hole,1} = \pi \left(\frac{D_1}{2}\right)^2 = \pi \left(\frac{0.6 \text{ m}}{2}\right)^2 = \pi (0.3 \text{ m})^2 = 0.09\pi \text{ m}^2 Ahole,1≈0.28274 m2A_{hole,1} \approx 0.28274 \text{ m}^2

  13. Calculate the area expansion coefficient (β\beta).

    β=2α=2×(12.5×10−6 /K)=25×10−6 /K\beta = 2\alpha = 2 \times (12.5 \times 10^{-6} \text{ /K}) = 25 \times 10^{-6} \text{ /K}

  14. Use the formula for area thermal expansion to find the increase in area (ΔAhole\Delta A_{hole}).

    ΔAhole=Ahole,1βΔT\Delta A_{hole} = A_{hole,1} \beta \Delta T ΔAhole=(0.09π m2)×(25×10−6 /K)×(400 K)\Delta A_{hole} = (0.09\pi \text{ m}^2) \times (25 \times 10^{-6} \text{ /K}) \times (400 \text{ K}) ΔAhole=0.09π m2×0.01\Delta A_{hole} = 0.09\pi \text{ m}^2 \times 0.01 ΔAhole=0.0009π m2\Delta A_{hole} = 0.0009\pi \text{ m}^2 ΔAhole≈0.002827 m2\Delta A_{hole} \approx 0.002827 \text{ m}^2

    The increase in area of the hole is 0.002827 m2\boxed{\text{0.002827 m}^2}.

    6. A piece of wire, original length 120 mm, is heated from 20 °C to 95 °C. If the final length of the wire is 120.324 mm, calculate the linear expansion coefficient of the wire.

  15. Identify the given values and calculate the change in temperature (ΔT\Delta T).

    Initial length, L1=120 mmL_1 = 120 \text{ mm}
    Final length, L2=120.324 mmL_2 = 120.324 \text{ mm}
    Initial temperature, T1=20 °CT_1 = 20 \text{ °C}
    Final temperature, T2=95 °CT_2 = 95 \text{ °C}

    ΔT=T2−T1=95 °C−20 °C=75 °C\Delta T = T_2 - T_1 = 95 \text{ °C} - 20 \text{ °C} = 75 \text{ °C}

  16. Use the linear thermal expansion formula and solve for the linear expansion coefficient (α\alpha).

    L2=L1(1+αΔT)L_2 = L_1 (1 + \alpha \Delta T) 120.324 mm=120 mm×(1+α×75 °C)120.324 \text{ mm} = 120 \text{ mm} \times (1 + \alpha \times 75 \text{ °C}) 120.324120=1+75α\frac{120.324}{120} = 1 + 75\alpha 1.0027=1+75α1.0027 = 1 + 75\alpha 0.0027=75α0.0027 = 75\alpha α=0.002775\alpha = \frac{0.0027}{75} α=3.6×10−5 /°C\alpha = 3.6 \times 10^{-5} \text{ /°C}

    The linear expansion coefficient of the wire is \boxed{\text{3.6 \times 10^{-5} /°C}}.

    7. The temperature of a right rectangular prism of carbon steel with dimensions 70×130×10 mm70 \times 130 \times 10 \text{ mm} is increased by 45 °C. The coefficient of linear expansion is 14×10−6 /°C14 \times 10^{-6} \text{ /°C}. Calculate the change in volume in mm3\text{mm}^3.

  17. Calculate the initial volume (V1V_1) of the prism.

    V1=70 mm×130 mm×10 mm=91 000 mm3V_1 = 70 \text{ mm} \times 130 \text{ mm} \times 10 \text{ mm} = 91\,000 \text{ mm}^3

  18. Identify the change in temperature (ΔT\Delta T) and the linear expansion coefficient (α\alpha).

    Calculate the cubic expansion coefficient (γ\gamma).

    ΔT=45 °C\Delta T = 45 \text{ °C} α=14×10−6 /°C\alpha = 14 \times 10^{-6} \text{ /°C} γ=3α=3×(14×10−6 /°C)=42×10−6 /°C\gamma = 3\alpha = 3 \times (14 \times 10^{-6} \text{ /°C}) = 42 \times 10^{-6} \text{ /°C}

  19. Use the formula for volumetric thermal expansion to find the change in volume (ΔV\Delta V).

    ΔV=V1γΔT\Delta V = V_1 \gamma \Delta T ΔV=(91 000 mm3)×(42×10−6 /°C)×(45 °C)\Delta V = (91\,000 \text{ mm}^3) \times (42 \times 10^{-6} \text{ /°C}) \times (45 \text{ °C}) ΔV=91 000 mm3×0.00189\Delta V = 91\,000 \text{ mm}^3 \times 0.00189 ΔV=172.0 mm3\Delta V = 172.0 \text{ mm}^3

    The change in volume is 172.0 mm3\boxed{\text{172.0 mm}^3}.

    8. The temperature of a steel rod with a diameter of 50 mm and a length of 150 mm is increased by 90 K. The increase in volume is 58 912.5 mm358\,912.5 \text{ mm}^3. Calculate:

    a) The percentage increase in volume.

  20. Calculate the initial volume (V1V_1) of the steel rod.

    V1=π(D2)2L=π(50 mm2)2×150 mmV_1 = \pi \left(\frac{D}{2}\right)^2 L = \pi \left(\frac{50 \text{ mm}}{2}\right)^2 \times 150 \text{ mm} V1=π(25 mm)2×150 mm=π×625 mm2×150 mmV_1 = \pi (25 \text{ mm})^2 \times 150 \text{ mm} = \pi \times 625 \text{ mm}^2 \times 150 \text{ mm} V1=93 750π mm3≈294 524.3 mm3V_1 = 93\,750\pi \text{ mm}^3 \approx 294\,524.3 \text{ mm}^3

  21. Calculate the percentage increase in volume.

    Percentage increase=ΔVV1×100%\text{Percentage increase} = \frac{\Delta V}{V_1} \times 100\% Percentage increase=58 912.5 mm3294 524.3 mm3×100%\text{Percentage increase} = \frac{58\,912.5 \text{ mm}^3}{294\,524.3 \text{ mm}^3} \times 100\% Percentage increase≈0.2000×100%=20.00%\text{Percentage increase} \approx 0.2000 \times 100\% = 20.00\%

    The percentage increase in volume is \boxed{\text{20.00%}}.

    b) The linear coefficient of expansion of steel.

  22. Use the formula for volumetric thermal expansion to find the cubic expansion coefficient (γ\gamma).

    ΔV=V1γΔT\Delta V = V_1 \gamma \Delta T 58 912.5 mm3=(93 750π mm3)×γ×(90 K)58\,912.5 \text{ mm}^3 = (93\,750\pi \text{ mm}^3) \times \gamma \times (90 \text{ K}) γ=58 912.593 750π×90\gamma = \frac{58\,912.5}{93\,750\pi \times 90} γ=58 912.526 507 189.4\gamma = \frac{58\,912.5}{26\,507\,189.4} γ≈0.0022225 /K\gamma \approx 0.0022225 \text{ /K}

  23. Calculate the linear expansion coefficient (α\alpha).

    α=γ3\alpha = \frac{\gamma}{3} α=0.0022225 /K3\alpha = \frac{0.0022225 \text{ /K}}{3} α≈7.408×10−4 /K\alpha \approx 7.408 \times 10^{-4} \text{ /K}

    The linear coefficient of expansion of steel is \boxed{\text{7.408 \times 10^{-4} /K}}.
    Self-correction: The percentage increase in volume is very high (20%), which leads to a very high linear expansion coefficient. Let's recheck the calculation for percentage increase in volume. If the increase in volume is 58912.5 mm^3, and the initial volume is 294524.3 mm^3, then the percentage increase is indeed 20%. This implies a very large expansion, which is unusual for typical materials. However, I must follow the given numbers.
    Let's re-evaluate the calculation for γ\gamma and α\alpha using the exact value of V1=93750π mm3V_1 = 93750\pi \text{ mm}^3 to avoid rounding errors.

    γ=58912.593750π×90=58912.58437500π\gamma = \frac{58912.5}{93750\pi \times 90} = \frac{58912.5}{8437500\pi} γ≈58912.526507189.438≈0.0022225 /K\gamma \approx \frac{58912.5}{26507189.438} \approx 0.0022225 \text{ /K} α=γ3≈0.00222253≈0.0007408 /K\alpha = \frac{\gamma}{3} \approx \frac{0.0022225}{3} \approx 0.0007408 \text{ /K}

    This value is indeed very high compared to typical linear expansion coefficients (which are usually in the order of 10−610^{-6}). However, based on the provided numbers, this is the correct calculation.

    9. A metal sphere has a volume of 1 m31 \text{ m}^3. The cubic coefficient of expansion is 36×10−6 /K36 \times 10^{-6} \text{ /K}. Calculate:

    a) The increase in volume if the temperature of the ball is increased by 200 K.

  24. Identify the given values.

    Initial volume, V1=1 m3V_1 = 1 \text{ m}^3
    Cubic expansion coefficient, γ=36×10−6 /K\gamma = 36 \times 10^{-6} \text{ /K}
    Change in temperature, ΔT=200 K\Delta T = 200 \text{ K}

  25. Use the formula for volumetric thermal expansion to find the increase in volume (ΔV\Delta V).

    ΔV=V1γΔT\Delta V = V_1 \gamma \Delta T ΔV=(1 m3)×(36×10−6 /K)×(200 K)\Delta V = (1 \text{ m}^3) \times (36 \times 10^{-6} \text{ /K}) \times (200 \text{ K}) ΔV=0.0072 m3\Delta V = 0.0072 \text{ m}^3

    The increase in volume is 0.0072 m3\boxed{\text{0.0072 m}^3}.

    b) The increase in diameter if the temperature is increased by 200 K.

  26. Calculate the initial diameter (D1D_1) of the sphere.

    V1=43π(D12)3V_1 = \frac{4}{3}\pi \left(\frac{D_1}{2}\right)^3 1 m3=43πD138=π6D131 \text{ m}^3 = \frac{4}{3}\pi \frac{D_1^3}{8} = \frac{\pi}{6} D_1^3 D13=6π m3D_1^3 = \frac{6}{\pi} \text{ m}^3 D1=6π3 m≈1.909863 m≈1.2407 mD_1 = \sqrt[3]{\frac{6}{\pi}} \text{ m} \approx \sqrt[3]{1.90986} \text{ m} \approx 1.2407 \text{ m}

  27. Calculate the linear expansion coefficient (α\alpha).

    α=γ3=36×10−6 /K3=12×10−6 /K\alpha = \frac{\gamma}{3} = \frac{36 \times 10^{-6} \text{ /K}}{3} = 12 \times 10^{-6} \text{ /K}

  28. Use the formula for linear thermal expansion to find the increase in diameter (ΔD\Delta D).

    ΔD=D1αΔT\Delta D = D_1 \alpha \Delta T ΔD=(1.2407 m)×(12×10−6 /K)×(200 K)\Delta D = (1.2407 \text{ m}) \times (12 \times 10^{-6} \text{ /K}) \times (200 \text{ K}) ΔD=1.2407 m×0.0024\Delta D = 1.2407 \text{ m} \times 0.0024 ΔD≈0.0029777 m\Delta D \approx 0.0029777 \text{ m}

    The increase in diameter is 0.002978 m\boxed{\text{0.002978 m}}.

    10. A metal sphere has a volume of 1 m31 \text{ m}^3. The temperature is increased by 60 °C. The linear expansion coefficient is 17×10−6 /°C17 \times 10^{-6} \text{ /°C}. Calculate:

    a) The increase in volume.

  29. Identify the given values.

    Initial volume, V1=1 m3V_1 = 1 \text{ m}^3
    Change in temperature, ΔT=60 °C\Delta T = 60 \text{ °C}
    Linear expansion coefficient, α=17×10−6 /°C\alpha = 17 \times 10^{-6} \text{ /°C}

  30. Calculate the cubic expansion coefficient (γ\gamma).

    γ=3α=3×(17×10−6 /°C)=51×10−6 /°C\gamma = 3\alpha = 3 \times (17 \times 10^{-6} \text{ /°C}) = 51 \times 10^{-6} \text{ /°C}

  31. Use the formula for volumetric thermal expansion to find the increase in volume (ΔV\Delta V).

    ΔV=V1γΔT\Delta V = V_1 \gamma \Delta T ΔV=(1 m3)×(51×10−6 /°C)×(60 °C)\Delta V = (1 \text{ m}^3) \times (51 \times 10^{-6} \text{ /°C}) \times (60 \text{ °C}) ΔV=0.00306 m3\Delta V = 0.00306 \text{ m}^3

    The increase in volume is 0.00306 m3\boxed{\text{0.00306 m}^3}.

    b) The final volume.

  32. Calculate the final volume (V2V_2) by adding the increase in volume to the initial volume.

    V2=V1+ΔVV_2 = V_1 + \Delta V V2=1 m3+0.00306 m3V_2 = 1 \text{ m}^3 + 0.00306 \text{ m}^3 V2=1.00306 m3V_2 = 1.00306 \text{ m}^3

    The final volume is 1.00306 m3\boxed{\text{1.00306 m}^3}.

    c) The final diameter.

  33. Calculate the initial diameter (D1D_1) of the sphere.

    V1=43π(D12)3V_1 = \frac{4}{3}\pi \left(\frac{D_1}{2}\right)^3 1 m3=π6D131 \text{ m}^3 = \frac{\pi}{6} D_1^3 D1=6π3 m≈1.2407 mD_1 = \sqrt[3]{\frac{6}{\pi}} \text{ m} \approx 1.2407 \text{ m}

  34. Use the linear thermal expansion formula to find the final diameter (D2D_2).

    D2=D1(1+αΔT)D_2 = D_1 (1 + \alpha \Delta T) D2=(1.2407 m)×(1+(17×10−6 /°C)×(60 °C))D_2 = (1.2407 \text{ m}) \times (1 + (17 \times 10^{-6} \text{ /°C}) \times (60 \text{ °C})) D2=1.2407 m×(1+0.00102)D_2 = 1.2407 \text{ m} \times (1 + 0.00102) D2=1.2407 m×1.00102D_2 = 1.2407 \text{ m} \times 1.00102 D2≈1.24196 mD_2 \approx 1.24196 \text{ m}

    The final diameter is 1.24196 m\boxed{\text{1.24196 m}}.

    11. A right rectangular prism of cast iron has dimensions of 50×40×30 mm50 \times 40 \times 30 \text{ mm} at 22 °C. The prism is heated to 85 °C. The coefficient of linear expansion of cast iron is 9×10−6 /°C9 \times 10^{-6} \text{ /°C}. Calculate:

    a) The increase in volume.

  35. Calculate the initial volume (V1V_1) of the prism.

    V1=50 mm×40 mm×30 mm=60 000 mm3V_1 = 50 \text{ mm} \times 40 \text{ mm} \times 30 \text{ mm} = 60\,000 \text{ mm}^3

  36. Identify the initial temperature, final temperature, and linear expansion coefficient.

    Calculate the change in temperature (ΔT\Delta T) and the cubic expansion coefficient (γ\gamma).
    Initial temperature, T1=22 °CT_1 = 22 \text{ °C}
    Final temperature, T2=85 °CT_2 = 85 \text{ °C}

    ΔT=T2−T1=85 °C−22 °C=63 °C\Delta T = T_2 - T_1 = 85 \text{ °C} - 22 \text{ °C} = 63 \text{ °C}

    Linear expansion coefficient, α=9×10−6 /°C\alpha = 9 \times 10^{-6} \text{ /°C}

    γ=3α=3×(9×10−6 /°C)=27×10−6 /°C\gamma = 3\alpha = 3 \times (9 \times 10^{-6} \text{ /°C}) = 27 \times 10^{-6} \text{ /°C}

  37. Use the formula for volumetric thermal expansion to find the increase in volume (ΔV\Delta V).

    ΔV=V1γΔT\Delta V = V_1 \gamma \Delta T ΔV=(60 000 mm3)×(27×10−6 /°C)×(63 °C)\Delta V = (60\,000 \text{ mm}^3) \times (27 \times 10^{-6} \text{ /°C}) \times (63 \text{ °C}) ΔV=60 000 mm3×0.001701\Delta V = 60\,000 \text{ mm}^3 \times 0.001701 ΔV=102.06 mm3\Delta V = 102.06 \text{ mm}^3

    The increase in volume is 102.06 mm3\boxed{\text{102.06 mm}^3}.

    b) The volume at 85 °C.

  38. Calculate the final volume (V2V_2) by adding the increase in volume to the initial volume.

    V2=V1+ΔVV_2 = V_1 + \Delta V V2=60 000 mm3+102.06 mm3V_2 = 60\,000 \text{ mm}^3 + 102.06 \text{ mm}^3 V2=60 102.06 mm3V_2 = 60\,102.06 \text{ mm}^3

    The volume at 85 °C is 60102.06 mm3\boxed{\text{60102.06 mm}^3}.

    12. A steel ball has a diameter of 60 mm and is 0.01 mm too large to pass through a hole in a brass plate when the ball and the plate are at a temperature of 303 K. Calculate the common temperature for the ball and plate at which the ball will just pass through the hole. The linear expansion of steel is 12×10−6 /K12 \times 10^{-6} \text{ /K} and that of brass 19×10−6 /K19 \times 10^{-6} \text{ /K}.

  39. Identify the initial diameters and expansion coefficients.

    Initial diameter of steel ball, Dsteel,1=60 mmD_{steel,1} = 60 \text{ mm}
    Initial diameter of brass hole, Dbrass,1=60 mm−0.01 mm=59.99 mmD_{brass,1} = 60 \text{ mm} - 0.01 \text{ mm} = 59.99 \text{ mm}
    Initial temperature, T1=303 KT_1 = 303 \text{ K}
    Linear expansion coefficient of steel, αsteel=12×10−6 /K\alpha_{steel} = 12 \times 10^{-6} \text{ /K}
    Linear expansion coefficient of brass, αbrass=19×10−6 /K\alpha_{brass} = 19 \times 10^{-6} \text{ /K}

  40. Set up the equation for the final diameters to be equal.

    Let T2T_2 be the final common temperature.

    Dsteel,2=Dbrass,2D_{steel,2} = D_{brass,2} Dsteel,1(1+αsteel(T2−T1))=Dbrass,1(1+αbrass(T2−T1))D_{steel,1} (1 + \alpha_{steel} (T_2 - T_1)) = D_{brass,1} (1 + \alpha_{brass} (T_2 - T_1))

    Let ΔT=T2−T1\Delta T = T_2 - T_1.

    Dsteel,1(1+αsteelΔT)=Dbrass,1(1+αbrassΔT)D_{steel,1} (1 + \alpha_{steel} \Delta T) = D_{brass,1} (1 + \alpha_{brass} \Delta T) Dsteel,1+Dsteel,1αsteelΔT=Dbrass,1+Dbrass,1αbrassΔTD_{steel,1} + D_{steel,1} \alpha_{steel} \Delta T = D_{brass,1} + D_{brass,1} \alpha_{brass} \Delta T Dsteel,1−Dbrass,1=Dbrass,1αbrassΔT−Dsteel,1αsteelΔTD_{steel,1} - D_{brass,1} = D_{brass,1} \alpha_{brass} \Delta T - D_{steel,1} \alpha_{steel} \Delta T Dsteel,1−Dbrass,1=(Dbrass,1αbrass−Dsteel,1αsteel)ΔTD_{steel,1} - D_{brass,1} = (D_{brass,1} \alpha_{brass} - D_{steel,1} \alpha_{steel}) \Delta T

  41. Solve for ΔT\Delta T.

    ΔT=Dsteel,1−Dbrass,1Dbrass,1αbrass−Dsteel,1αsteel\Delta T = \frac{D_{steel,1} - D_{brass,1}}{D_{brass,1} \alpha_{brass} - D_{steel,1} \alpha_{steel}} ΔT=60 mm−59.99 mm(59.99 mm×19×10−6 /K)−(60 mm×12×10−6 /K)\Delta T = \frac{60 \text{ mm} - 59.99 \text{ mm}}{(59.99 \text{ mm} \times 19 \times 10^{-6} \text{ /K}) - (60 \text{ mm} \times 12 \times 10^{-6} \text{ /K})} ΔT=0.01 mm(1139.81×10−6 mm/K)−(720×10−6 mm/K)\Delta T = \frac{0.01 \text{ mm}}{(1139.81 \times 10^{-6} \text{ mm/K}) - (720 \times 10^{-6} \text{ mm/K})} ΔT=0.01 mm(1139.81−720)×10−6 mm/K\Delta T = \frac{0.01 \text{ mm}}{(1139.81 - 720) \times 10^{-6} \text{ mm/K}} ΔT=0.01419.81×10−6 K\Delta T = \frac{0.01}{419.81 \times 10^{-6}} \text{ K} ΔT≈23.82 K\Delta T \approx 23.82 \text{ K}

  42. Calculate the final temperature (T2T_2).

    T2=T1+ΔTT_2 = T_1 + \Delta T T2=303 K+23.82 KT_2 = 303 \text{ K} + 23.82 \text{ K} T2=326.82 KT_2 = 326.82 \text{ K}

    The common temperature at which the ball will just pass through the hole is 326.82 K\boxed{\text{326.82 K}}.

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