91影视

• The coefficient of thermal expansion of brass is\({\alpha _{{\rm{brass}}}} = 19 \times {10^{ - 6}}\;{\rm{/^\circ C}}\).
• The coefficient of thermal expansion of iron is\({\alpha _{{\rm{iron}}}} = 12 \times {10^{ - 6}}\;{\rm{/^\circ C}}\).
• The initial diameter of the ring is\({d_{{\rm{iRing}}}} = 8.741{\rm{ cm}}\).
• The initial diameter of the hole or plug is\({d_{\rm{i}}}_{{\rm{Hole}}} = 8.755{\rm{ cm}}\).

The initial temperature of the hole and plug is \({T_{\rm{i}}} = 15^\circ {\rm{C}}\).

The final diameter of the ring and the hole is the same. The final diameter of the ring or hole is the sum of its initial diameter and change in the diameter.

In this problem, equate the sum of diameters and the change in the diameters of the ring and hole. From these equations, the common temperature of both can be evaluated.

The ring is made up of iron, and the hole is made up of brass. The final volume of the ring and hole can be expressed as shown below:

\(\begin{aligned}{c}{\left( {{d_{\rm{i}}} + \Delta d} \right)_{{\rm{Ring}}}} &= {\left( {{d_{\rm{i}}} + \Delta d} \right)_{{\rm{Hole}}}}\\{d_{{\rm{iRing}}}} + \Delta {d_{{\rm{Ring}}}} &= {d_{\rm{i}}}_{{\rm{Hole}}} + \Delta {d_{{\rm{Hole}}}}\\{d_{{\rm{iRing}}}} - {d_{\rm{i}}}_{{\rm{Hole}}} &= \Delta {d_{{\rm{Hole}}}} - \Delta {d_{{\rm{Ring}}}}\\{d_{{\rm{iRing}}}} - {d_{\rm{i}}}_{{\rm{Hole}}} &= {\alpha _{{\rm{brass}}}}{d_{{\rm{iHole}}}}\Delta T - {\alpha _{{\rm{iron}}}}{d_{{\rm{iRing}}}}\Delta T\end{aligned}\)

This can be further solved as shown below:

\(\begin{aligned}{c}{d_{{\rm{iRing}}}} - {d_{\rm{i}}}_{{\rm{Hole}}} &= \Delta T\left( {{\alpha _{{\rm{brass}}}}{d_{{\rm{iHole}}}} - {\alpha _{{\rm{iron}}}}{d_{{\rm{iRing}}}}} \right)\\\Delta T &= \frac{{{d_{{\rm{iRing}}}} - {d_{\rm{i}}}_{{\rm{Hole}}}}}{{\left( {{\alpha _{{\rm{brass}}}}{d_{{\rm{iHole}}}} - {\alpha _{{\rm{iron}}}}{d_{{\rm{iRing}}}}} \right)}}\\\left( {{T_{\rm{f}}} - {T_{\rm{i}}}} \right) &= \frac{{{d_{\rm{i}}}_{{\rm{Hole}}} - {d_{{\rm{iRing}}}}}}{{\left( {{\alpha _{{\rm{iron}}}}{d_{{\rm{iRing}}}} - {\alpha _{{\rm{brass}}}}{d_{{\rm{iHole}}}}} \right)}}\\{T_{\rm{f}}} &= {T_{\rm{i}}} + \frac{{{d_{\rm{i}}}_{{\rm{Hole}}} - {d_{{\rm{iRing}}}}}}{{\left( {{\alpha _{{\rm{iron}}}}{d_{{\rm{iRing}}}} - {\alpha _{{\rm{brass}}}}{d_{{\rm{iHole}}}}} \right)}}\end{aligned}\)

Here,\({T_{\rm{f}}}\)is the final or common temperature of the ring and the plug to be fit.

Substitute the values in the above equation.

\(\begin{aligned}{c}{T_{\rm{f}}} &= 15^\circ {\rm{C}} + \frac{{\left( {8.755{\rm{ cm}} - 8.741{\rm{ cm}}} \right)}}{{\left( {12 \times {{10}^{ - 6}}\;{\rm{/^\circ C}} \times 8.741{\rm{ cm}} - 19 \times {{10}^{ - 6}}\;{\rm{/^\circ C}} \times 8.755{\rm{ cm}}} \right)}}\\ &= 15^\circ {\rm{C}} + \left( {\frac{{0.01{\rm{ cm}}}}{{ - 6.145 \times {{10}^{ - 5}}\;{\rm{cm/^\circ C}}}}} \right)\\ &= 15^\circ {\rm{C}} + \left( { - 227.83{\rm{^\circ C}}} \right)\\ &= - 212.83{\rm{^\circ C}}\end{aligned}\)

Thus, the temperature of the ring and the plug to be fit is \(\left( { - 212.83{\rm{^\circ C}}} \right)\).