Question

at low temperature, the specific heats of solids are typically proportional to (t^{3}). the first understanding of this behavior was due to the dutch physicist peter debye, who in 1912, treated atomic oscillations with the quantum theory that max planck had recently used for radiation. for instance, a good approximation for the specific heat of salt, (nacl) is (c = 3.33\times10^{4} j/(kgcdot k) (t/321 k)^{3}). the constant is called debye temperature of (nacl) and the formula works well when (t<0.04 theta_{d}). using the formula, how much heat is required to raise the temperature of (m = 240 g) of (nacl) from (t_{i}=5.0 k) to (t_{f}=15.0 k)? (delta q
ightarrow dt
ightarrow dq = mc dt), (varphi=int dq=int_{t_{i}}^{t_{f}}mc dt), (varphi = mint_{t_{i}}^{t_{f}}c dt=mint_{t_{i}}^{t_{f}}3.33\times10^{3}\frac{j}{kgcdot k}\frac{t^{3}}{(321 k)^{3}}dt), (varphi=\frac{0.240 kgcdot3.33\times10^{3} j}{321^{3}cdot k^{3}}int_{5.0}^{15.0}t^{3} dt), (varphi = 0.240\times3.33\times10^{3}\frac{1}{321^{3}}left\frac{t^{4}}{4}
ight_{5.0}^{15.0} j), (varphi =? j)

Explanation:

Step1: Convert mass to SI - unit

The mass of NaCl is $m = 240\ g=0.24\ kg$.

Step2: Identify the specific - heat formula

The specific - heat formula is $c = 3.33\times10^{4}\frac{J}{kg\cdot K}(\frac{T}{321\ K})^{3}$.

Step3: Set up the heat - transfer integral

The heat transfer formula is $Q=\int_{T_1}^{T_2}mc\ dT$. Substitute $m$ and $c$ into the formula:
\[

$$\begin{align*} Q&=m\int_{T_1}^{T_2}3.33\times 10^{4}\frac{J}{kg\cdot K}(\frac{T}{321\ K})^{3}dT\\ &=m\times3.33\times 10^{4}\frac{J}{kg\cdot K}\times\frac{1}{(321\ K)^{3}}\int_{T_1}^{T_2}T^{3}dT \end{align*}$$

\]

Step4: Integrate $T^{3}$

We know that $\int T^{3}dT=\frac{1}{4}T^{4}+C$. Evaluating the definite integral $\int_{T_1}^{T_2}T^{3}dT=\frac{1}{4}(T_2^{4}-T_1^{4})$. Here, $T_1 = 5.0\ K$ and $T_2 = 15.0\ K$.

Step5: Calculate the heat

\[

$$\begin{align*} Q&=0.24\ kg\times3.33\times 10^{4}\frac{J}{kg\cdot K}\times\frac{1}{(321\ K)^{3}}\times\frac{1}{4}[(15.0\ K)^{4}-(5.0\ K)^{4}]\\ &=0.24\times3.33\times 10^{4}\times\frac{1}{321^{3}}\times\frac{1}{4}(15^{4}-5^{4})\\ &=0.24\times3.33\times 10^{4}\times\frac{1}{321^{3}}\times\frac{1}{4}(50625 - 625)\\ &=0.24\times3.33\times 10^{4}\times\frac{1}{321^{3}}\times\frac{1}{4}\times50000 \end{align*}$$

\]
\[

$$\begin{align*} 321^{3}&=321\times321\times321 = 33076161\\ Q&=0.24\times3.33\times 10^{4}\times\frac{1}{33076161}\times\frac{1}{4}\times50000\\ &=0.24\times3.33\times 10^{4}\times\frac{50000}{4\times33076161}\\ &=0.24\times3.33\times 10^{4}\times\frac{12500}{33076161}\\ &=0.24\times3.33\times\frac{12500\times10^{4}}{33076161}\\ &=0.24\times3.33\times\frac{125000000}{33076161}\\ &=0.24\times3.33\times 3.78\\ &=0.24\times 12.53\\ & = 3.01\ J \end{align*}$$

\]

Answer:

$3.01\ J$