During the construction of a bridge, an engineer must insert a steel support cable through a pre-drilled hole into a concrete support beam. The cable has a diameter of 7 cm and a mass of 40 kg, but the hole has a diameter of only 6.99 cm. Take the steel linear expansion coefficient to be \(12\times10^{- 6} \,\text{K}^{-1}\) and the specific heat of steel to be 460 J /kg K. Assume both the cable and the beam are at room temperature.

a) How could the engineer adjust the temperature of the cable so that it \({just}\) fits into the hole?

b) How much thermal energy must the steel lose for this to happen?

a) Use the equation for Linear Thermal Expansion to solve for the change in temperature.

b) Find the heat, given the change in temperature already found in part (a).

a) Linear thermal expansion can be written as:

\begin{equation*}
\Delta D =  \alpha D \Delta T,
\end{equation*}

where solving for \( \Delta T\) and replacing the numerical values we obtain:

\begin{equation*}
\Delta T = -119.04 \, ^{\circ} \text{C}.
\end{equation*}

b) The heat for the steel can be calculated as:

\begin{equation*}
Q = m c \Delta T,
\end{equation*}

where \( \Delta T\) was found in a). Then:

\begin{equation*}
Q = -2.19 \times 10^6 \ \text{J}.
\end{equation*}

For a more detailed explanation of any of these steps, click on “Detailed Solution”.

a) We must use the coefficient of linear expansion of steel to relate the temperature change with the change in diameter required for the cable to fit into the hole.

As the temperature of the cable decreases, its diameter \(D\) will reduce. The fractional change in diameter \(\frac{\Delta D}{D}\) is directly proportional to the temperature change \(\Delta T\). We can write this as

\begin{equation}
\frac{\Delta D}{D} = \alpha \Delta T,
\end{equation}

where the proportionality constant \(\alpha\) is the coefficient of linear expansion of steel, which describes how much length changes for a given temperature change. We can rewrite this equation as

\begin{equation}
\frac{D_f – D_i}{D_i} = \alpha \Delta T,
\end{equation}

where \(D_f\) and \(D_i\) are the diameters of the cable after and before the temperature change, respectively. Dividing by \(\alpha\) gives

\begin{equation}
\Delta T = \frac{D_f – D_i}{\alpha D_i}.
\end{equation}

In order for the cable to barely fit, its final diameter \(D_f\) must be equal to the diameter of the hole. If we let \(D_h\) be the diameter of the hole, we obtain that the temperature change is given by

\begin{equation}
\label{EQ:DT} \Delta T = \frac{D_h – D_i}{\alpha D_i},
\end{equation}

and inserting numerical values yields

\begin{equation}
\Delta T = \frac{ 6.99 \ \text{cm} – 7 \ \text{cm}}{(12 \times 10^{-6} \text{C}^{-1})\cdot (7 \ \text{cm})}
= -119.04 \, ^{\circ} \text{C},
\end{equation}

where the negative sign indicates that the temperature decreases.

b) We should use the specific heat of steel to relate its temperature change with the heat exchanged by it.

The following equation relates the heat transfer between the steel cable and its surroundings to the change in temperature \(\Delta T\):

\begin{equation}
Q = m c \Delta T,
\end{equation}

where \(m\) is the mass of the cable and \(c\) is the specific heat of steel, which measures how much heat is required to be transferred per unit mass to achieve a given temperature change. If we substitute eq. \eqref{EQ:DT}, we get

\begin{equation}
Q = m c \frac{D_h – D_i}{\alpha D_i},
\end{equation}

and inserting numerical values gives

\begin{equation}
Q = \left(40 \ \text{kg}\right) \left(460 \ \frac{\text{J}}{\ \text{kg} \, ^{\circ}\text{C}} \right) \frac{6.99 \ \text{cm} – 7 \ \text{cm}}{ \left(12 \times 10^{-6} \, ^{\circ}\text{C}^{-1}\right)\left(7 \ \text{cm}\right)} = -2.19 \times 10^6 \ \text{J}.
\end{equation}

Note that the heat is negative, as required since the heat exits from the steel cable.

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