A 2.00-kg copper rod is 50.00 cm long at 23 $\degree$C.If 40 000 J are transferred to the rod by heat,what is its change in length? c_copper = 387 J/kg*$\degree$C and a_copper = 17 ´ 10^-6/ $\degree$C.

Question

Quizplus · Accepted Answer

First, we need to calculate the change in temperature of the rod from 23ºC to the final temperature. Let's assume the final temperature is T. We can use the formula Q = mcΔT, where Q is the heat transferred, m is the mass of the rod, c is the specific heat capacity of copper, and ΔT is the change in temperature.

So, ΔT = Q/(mc) = (40000 J)/(2.00 kg × 387 J/kgºC) = 51.68ºC

Now, we can use the linear expansion formula ΔL = αLΔT, where ΔL is the change in length, α is the coefficient of linear expansion for copper, L is the original length of the rod, and ΔT is the change in temperature.

ΔL = αLΔT = (17 × 10^-6 / ºC) × (50.00 cm) × (51.68ºC) = 0.044 cm

Therefore, the answer is B) 0.044 cm.

A 200-Kg Copper Rod Is 50 °\degree° CIf 40 000 J Are Transferred to the Rod by \degree

A 200-Kg Copper Rod Is 50 $\degree$ CIf 40 000 J Are Transferred to the Rod by \degree