Question

In an insulated container an 100 W electric heating element of small mass warms up a 300 g sample of copper for 6 s. The initial temperature of the copper was $\mathbf{20}\mathbf{°}$(roomtemperature). Predict the final temperature of the copper, using the 3kB specific heat per atom.

Short answer

The final temperature of copper is 317 K.

Step-by-step solution

Identification of given data

The given data can be listed below,

• The mass of sample of copper is, m=300 g.
• The time for which copper sample warm up, t =6 s .
• The initial temperature of the copper sample is, $T=20°C=293\mathrm{K}$.

The power dissipated by heating element is,$P=100W$

Concept/Significance of specific heat capacity

The amount of heat required to increase the temperature of a gramme of material by one unit.

Determination of the final temperature of the copper, using the 3kB specific heat per atom

The heat transferred by heating element is given by,

Q=Pt

Here, P is the power dissipated by heating element and t is the time.

Substitute values in the above,

$$\begin{gathered} \mathrm{Q}=100\mathrm{W}\times 6\mathrm{s} \\ =600\mathrm{J} \end{gathered}$$

The heat transferred to copper is given by,

$Q_{cu}=c\left(T_f-T_i\right)$

Here,c is specific heat capacity whose value is $3k_B{N}_A$,$T_f$is the final temperature and is the initial temperature of copper.

The final temperature of the copper sample is given by,

$T_f=T_i+\frac{Q}{3k_B{N}_A}$

Substitute all the values in the above equation.

${\mathrm{T}}_{\mathrm{f}}=293\mathrm{K}+\frac{600\mathrm{W}}{3\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{mol}.\mathrm{K}\right)\left(6.023\times {10}^{-23}{\mathrm{mol}}^{-1}\right)}$

$=317\mathrm{K}$

Thus, the final temperature of copper is 317 K .