The coefficient of linear expansion of copper is $17 \times 10 ^ { - 6 } K ^ { - 1 }$ A block of copper 30 cm wide, 45 cm Iong, and 10 cm thick is heated from $0 ^ { \circ } \mathrm { C } \text { to } 100 ^ { \circ } \mathrm { C }$
What is the change in the volume of the block?
A) $6.9 \times 10 ^ { - 5 } \mathrm {~m} ^ { 3 }$
B) $4.6 \times 10 ^ { - 5 } \mathrm {~m} ^ { 3 }$
C) $14 \times 10 ^ { - 5 } m ^ { 3 }$
D) $2.3 \times 10 ^ { - 5 } \mathrm {~m} ^ { 3 }$
E) $5.2 \times 10 ^ { - 5 } \mathrm {~m} ^ { 3 }$

Question

The coefficient of linear expansion of copper is $17 \times 10 ^ { - 6 } K ^ { - 1 }$ A block of copper  30 cm wide,  45 cm Iong, and  10 cm thick is heated from $0 ^ { \circ } \mathrm { C } \text { to } 100 ^ { \circ } \mathrm { C }$
What is the change in the volume of the block?
A) $6.9 \times 10 ^ { - 5 } \mathrm {~m} ^ { 3 }$
B) $4.6 \times 10 ^ { - 5 } \mathrm {~m} ^ { 3 }$
C) $14 \times 10 ^ { - 5 } m ^ { 3 }$
D) $2.3 \times 10 ^ { - 5 } \mathrm {~m} ^ { 3 }$
E) $5.2 \times 10 ^ { - 5 } \mathrm {~m} ^ { 3 }$

Quizplus · Accepted Answer

The change in volume ($\Delta V$) due to thermal expansion can be calculated using the formula $\Delta V = \beta V_0 \Delta T$, where $\beta$ is the volumetric expansion coefficient, $V_0$ is the initial volume, and $\Delta T$ is the change in temperature. For solids, $\beta \approx 3\alpha$, where $\alpha$ is the coefficient of linear expansion. Given $\alpha = 17 \times 10^{-6} K^{-1}$, $\beta = 3 \times 17 \times 10^{-6} K^{-1} = 51 \times 10^{-6} K^{-1}$. The initial volume $V_0 = 0.3 m \times 0.45 m \times 0.1 m = 0.0135 m^3$, and $\Delta T = 100^\circ C$. Thus, $\Delta V = 51 \times 10^{-6} K^{-1} \times 0.0135 m^3 \times 100 K = 6.885 \times 10^{-5} m^3$, which rounds to $6.9 \times 10^{-5} m^3$.

The Coefficient of Linear Expansion of Copper Is 17×10−6K−117 \times 10 ^ { - 6 } K ^ { - 1 }17×10−6K−1

The Coefficient of Linear Expansion of Copper Is $17 \times 10 ^ { - 6 } K ^ { - 1 }$