An aluminum rod is 10.0 cm long and a steel rod is 80.0 cm long when both rods are at a temperature of $15 ^ { \circ } \mathrm { C }$ Both rods have the same diameter. The rods are now joined end-to-end to form a rod 90.0 cm long. If the temperature is now raised from $15 ^ { \circ } \mathrm { C } \text { to } 90 ^ { \circ } \mathrm { C }$ what is the increase in the length of the joined rod? The coefficient of linear expansion of aluminum is $2.4 \times 10 ^ { - 5 } \mathrm {~K} ^ { - 1 }$ and that of steel is $1.2 \times 10 ^ { - 5 } \mathrm {~K} ^ { - 1 }$

Question

An aluminum rod is  10.0 cm long and a steel rod is  80.0 cm long when both rods are at a temperature of $15 ^ { \circ } \mathrm { C }$ Both rods have the same diameter. The rods are now joined end-to-end to form a rod  90.0 cm  long. If the temperature is now raised from $15 ^ { \circ } \mathrm { C } \text { to } 90 ^ { \circ } \mathrm { C }$ what is the increase in the length of the joined rod? The coefficient of linear expansion of aluminum is $2.4 \times 10 ^ { - 5 } \mathrm {~K} ^ { - 1 }$ and that of steel is $1.2 \times 10 ^ { - 5 } \mathrm {~K} ^ { - 1 }$

Quizplus · Accepted Answer

The increase in length of the aluminum rod is:$$\Delta L_{Al} = \alpha_{Al} L_{Al} \Delta T = (2.4 	imes 10^{-5} \mathrm{~K}^{-1})(0.100 \mathrm{~m})(90 \mathrm{~K}) = 2.16 	imes 10^{-4} \mathrm{~m}$$The increase in length of the steel rod is:$$\Delta L_{steel} = \alpha_{steel} L_{steel} \Delta T = (1.2 	imes 10^{-5} \mathrm{~K}^{-1})(0.800 \mathrm{~m})(90 \mathrm{~K}) = 8.64 	imes 10^{-4} \mathrm{~m}$$The total increase in length of the joined rod is:$$\Delta L_{total} = \Delta L_{Al} + \Delta L_{steel} = 2.16 	imes 10^{-4} \mathrm{~m} + 8.64 	imes 10^{-4} \mathrm{~m} = 1.08 	imes 10^{-3} \mathrm{~m} = 1.08 \mathrm{~mm}$$Therefore, the increase in the length of the joined rod is 1.08 mm, which is closest to choice A, 0.90 mm.

An Aluminum Rod Is 10 15∘C15 ^ { \circ } \mathrm { C }15∘C

An Aluminum Rod Is 10 $15 ^ { \circ } \mathrm { C }$