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An iron tyre with an initial diameter of 60 cm is to be fitted onto a steel wheel of the same diameter. If the temperature of the tyre is increased to 645°C, what is the new diameter of the tyre? The

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Answer

645∘C\text{645}^\circ\text{C}

Step-by-step solution

4 steps

  1. Identify the given values and the formula for linear expansion.

    The initial diameter is D0=60 cmD_0 = 60 \text{ cm}.
    The final diameter is D=60.45 cmD = 60.45 \text{ cm}.
    The initial temperature is T0=20∘CT_0 = 20^\circ\text{C}.
    The linear expansivity of iron is α=1.2×10−5 K−1\alpha = 1.2 \times 10^{-5} \text{ K}^{-1}.
    The formula for linear expansion is D=D0(1+αΔT)D = D_0 (1 + \alpha \Delta T), where ΔT=T−T0\Delta T = T - T_0.

  2. Rearrange the formula to solve for the change in temperature, ΔT\Delta T.

    D=D0(1+αΔT)D = D_0 (1 + \alpha \Delta T) DD0=1+αΔT\frac{D}{D_0} = 1 + \alpha \Delta T DD0−1=αΔT\frac{D}{D_0} - 1 = \alpha \Delta T ΔT=1α(DD0−1)\Delta T = \frac{1}{\alpha} \left( \frac{D}{D_0} - 1 \right)

  3. Substitute the given values into the rearranged formula and calculate ΔT\Delta T.

    ΔT=11.2×10−5 K−1(60.45 cm60 cm−1)\Delta T = \frac{1}{1.2 \times 10^{-5} \text{ K}^{-1}} \left( \frac{60.45 \text{ cm}}{60 \text{ cm}} - 1 \right) ΔT=11.2×10−5(1.0075−1)\Delta T = \frac{1}{1.2 \times 10^{-5}} (1.0075 - 1) ΔT=11.2×10−5(0.0075)\Delta T = \frac{1}{1.2 \times 10^{-5}} (0.0075) ΔT=0.00751.2×10−5\Delta T = \frac{0.0075}{1.2 \times 10^{-5}} ΔT=625 K\Delta T = 625 \text{ K}

    Since a change in temperature of 1 K1 \text{ K} is equal to a change of 1∘C1^\circ\text{C}, ΔT=625∘C\Delta T = 625^\circ\text{C}.

  4. Calculate the final temperature TT.

    T=T0+ΔTT = T_0 + \Delta T T=20∘C+625∘CT = 20^\circ\text{C} + 625^\circ\text{C} T=645∘CT = 645^\circ\text{C}

    The temperature to which the tyre must be heated is 645∘C\text{645}^\circ\text{C}.

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