Question

An aluminum rod and a nickel rodare both 5.00 m long at 20.0°C.The temperature of each is raisedto 70.0°C. What is the differencein length between the two rods?AluminumNickela = 23.10-6C+ B = 69.10-6 0-1a = 13.10-6C1 B = 39.10-6-1(Unit = m)Enter

Answer 1

Answer:

The difference in length between the two rods = 0.0025m

Explanations:

Linear expansivity of a material is given by the formula:

`\alpha\text{ = }(l_2-l_1)/(l_1(\theta_2-\theta_1))`

For the Aluminium rod:

`\begin{gathered} l_(A1)\text{ = 5.0m} \\ \theta_(A1)=20^0C \\ \theta_(A2)=70^0C \\ \alpha_A\text{ = }23*10^(-6)C^(-1) \\ \alpha_A\text{ = }(l_(A2)-l_(A1))/(l_(A1)(\theta_(A2)-\theta_(A1))) \\ \text{ }23*10^(-6)\text{ = }(l_(A2)-5)/(5(70-20)) \\ 5*50*\text{ }23*10^(-6)=\text{ }l_(A2)-5 \\ l_(A2)=\text{ (}5750\text{ }*10^(-6))\text{ + 5} \\ l_(A2)=\text{ 0.00575+5} \\ l_(A2)=\text{ 5.00575m} \end{gathered}`

For the Nickel rod:

`\begin{gathered} l_(N1)\text{ = 5.0m} \\ \theta_(N1)=20^0C \\ \theta_(N2)=70^0C \\ \alpha_N=\text{ 13}*10^(-6)C^(-1) \\ \alpha_N\text{ = }(l_(N2)-l_(N1))/(l_(N1)(\theta_(N2)-\theta_(N1))) \\ \text{ 1}3*10^(-6)\text{ = }(l_(N2)-5)/(5(70-20)) \\ 5*50*\text{ 1}3*10^(-6)=\text{ }l_(A2)-5 \\ l_(N2)=\text{ (32}50\text{ }*10^(-6))\text{ + 5} \\ l_(N2)=\text{ }0.00325+5 \\ l_(N2)=\text{ 5.00325m} \end{gathered}`

The difference in length between the two rods will be given as:

`\begin{gathered} l_(A2)-l_(N2)=\text{ 5.00575-5.00325} \\ l_(A2)-l_(N2)=0.0025m \end{gathered}`

The difference in length between the two rods = 0.0025m