Question

A 3 kg, 30 cm long iron rod is heated from 20ºC to 80ºC. Iron has an α of 1.2 × 10-5 ºC-1 and a c of 450 J/kgºC. a) How much did its length increase when heated up?- 0.2 mm- 2 mm- 2 cm- 20 cmb) how much heat was required for this expansion?- 810 kJ- 11 kJ- 81 kJ- 108 kJc) With the rod now at 80ºC, you want to cool it back down to 20ºC by dropping it into a bucket of ice water at 0ºC. How much water would be required?- 0.5 kg- 1 kg- 2.3 kg- 0.1 kg

Answer 1

Given data:

* The actual length of the iron rod is 30 cm.

* The initial temperature is 20 degree celsius.

* The final temperature is 80 degree celsius.

* The mass of the iron rod is 3 kg.

* The value of constants are,

$\begin{gathered} \alpha=1.2\times10^{-5\circ}C^{-1} \\ c=450Jkg^{-1\circ}C^{-1} \end{gathered}$

Solution:

(a). The increase in the length of the iron rod with the change in the temperature is,

$\alpha=\frac{dL}{LdT}$

where L is the actual length, and dL is the change in the length,

Substituting the known values,

$\begin{gathered} 1.2\times10^{-5}=\frac{dL^{}}{30\times10^{-2}\times(80^{\circ}-20^{\circ})} \\ 1.2\times10^{-5}=\frac{dL^{}}{1800\times10^{-2}} \\ 1.2\times10^{-5}=\frac{dL^{}}{18} \\ dL=1.2\times10^{-5}\times18 \\ dL=21.6\times10^{-5}\text{ m} \\ dL=0.2\times0^{-3}\text{ m} \\ dL=0.2\text{ mm} \end{gathered}$

Thus, the change in the length of the iron rod is 0.2 mm.

Hence, first option is the correct answer.