Answer to Question #209839 in Molecular Physics | Thermodynamics for Nkanyezi

Question #209839
A  brass  plug has  a  diameter  of 10,000 cm  at  20°C.  To what  temperature  must  it  be  cooled in order to fit  into  a  hole  A  brass  plug has  a  diameter  of 10,000 cm  at  20°C.  To what  temperature  must  it  be  cooled in order to fit  into  a  hole  with a  diameter  of 9,997 cm?  (α  =19×  106)





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Expert's answer
2021-06-24T11:58:10-0400

By the definition of the linear thermal expansion we have:



ΔL=αL0ΔT,\Delta L = \alpha L_0 \Delta T,

here, ΔL=LL0\Delta L = L - L_0 is the increace in diameter of the brass plug when it is heated, L0=10 mL_0 = 10 \ m is the initial diameter of the brass plug at some temperature T0T_0L=10 mL = 10 \ m is the diameter of the brass plug at T=20 CT = 20 \ ^{\circ}Cα=19106 C1\alpha = 19 \cdot 10^{-6} \ ^{\circ}C^{-1} is the linear expansion coefficient for steel, ΔT=TT0\Delta T = T - T_0 is the change in temperature.

From this formula we can find the change in temperature:



ΔT=LL0αL0,\Delta T = \dfrac{L - L_0}{\alpha L_0},ΔT=10 m9.997 m19106 C19.997 m=15.79 C.\Delta T = \dfrac{10 \ m - 9.997 \ m}{19 \cdot 10^{-6} \ ^{\circ}C^{-1} \cdot 9.997\ m} = 15.79 \ ^{\circ}C.

Finally, we can calculate the temperature at which the brass plug will fit exactly into a hole of constant diameter:



T0=TΔT=20 C15.79 C=4.21 C.T_0 = T - \Delta T = 20 \ ^{\circ}C - 15.79 \ ^{\circ}C = 4.21 \ ^{\circ}C.

Answer:

T0=4.21 C.T_0 = 4.21 \ ^{\circ}C.

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