A piece of metal weighs 46 , g in air. When i | vedclass

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(B) Loss of weight at $27^{\circ}C$ is $W_1 = 46 - 30 = 16 \, g$. This equals the weight of the displaced liquid: $V_1 	imes ho_1 	imes g = 16 \,

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$2.316 	imes 10^{-5} \, ^{\circ}C^{-1}$

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(B) Loss of weight at $27^{\circ}C$ is $W_1 = 46 - 30 = 16 \, g$. This equals the weight of the displaced liquid: $V_1 	imes ho_1 	imes g = 16 \, g$,where $ho_1 = 1.24 \, g/cm^3$. Thus,$V_1 = 16 / 1.24 \, cm^3$.Loss of weight at $42^{\circ}C$ is $W_2 = 46 - 30.5 = 15.5 \, g$. This equals $V_2 	imes ho_2 	imes g = 15.5 \, g$,where $ho_2 = 1.20 \, g/cm^3$. Thus,$V_2 = 15.5 / 1.20 \, cm^3$.The volume expansion of the metal is given by $V_2 = V_1(1 + \gamma \Delta T)$,where $\gamma = 3\alpha$ and $\Delta T = 42 - 27 = 15^{\circ}C$.$\frac{V_2}{V_1} = \frac{15.5 / 1.20}{16 / 1.24} = \frac{15.5 	imes 1.24}{16 	imes 1.20} = \frac{19.22}{19.2} \approx 1.0010416$.Since $1 + 3\alpha(15) = 1.0010416$,we have $3\alpha(15) = 0.0010416$.$\alpha = \frac{0.0010416}{45} \approx 2.315 	imes 10^{-5} \, ^{\circ}C^{-1}$.

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