To solve the problem of calculating the heat required to raise the temperature of 2 moles of rock salt from 10 K to 50 K using Debye's T³ law, we can follow these steps:
### Step 1: Write the expression for heat capacity
According to Debye's T³ law, the molar heat capacity \( C \) is given by:
\[
C = \frac{k T^3}{\theta^3}
\]
where \( k = 1940 \, \text{J mol}^{-1} \text{K}^{-1} \) and \( \theta = 281 \, \text{K} \).
### Step 2: Set up the heat transfer equation
The heat \( dQ \) required to change the temperature by an infinitesimal amount \( dT \) is given by:
\[
dQ = n C \, dT
\]
where \( n \) is the number of moles. For our case, \( n = 2 \) moles.
### Step 3: Substitute for \( C \)
Substituting the expression for \( C \) into the heat transfer equation:
\[
dQ = n \left( \frac{k T^3}{\theta^3} \right) dT
\]
Thus,
\[
dQ = 2 \left( \frac{1940 T^3}{281^3} \right) dT
\]
### Step 4: Integrate to find total heat \( Q \)
To find the total heat required to raise the temperature from \( T_1 = 10 \, \text{K} \) to \( T_2 = 50 \, \text{K} \), we integrate:
\[
Q = \int_{T_1}^{T_2} dQ = \int_{10}^{50} 2 \left( \frac{1940 T^3}{281^3} \right) dT
\]
This simplifies to:
\[
Q = \frac{2 \cdot 1940}{281^3} \int_{10}^{50} T^3 \, dT
\]
### Step 5: Calculate the integral
The integral of \( T^3 \) is:
\[
\int T^3 \, dT = \frac{T^4}{4}
\]
Thus,
\[
\int_{10}^{50} T^3 \, dT = \left[ \frac{T^4}{4} \right]_{10}^{50} = \frac{50^4}{4} - \frac{10^4}{4}
\]
Calculating \( 50^4 \) and \( 10^4 \):
\[
50^4 = 6250000 \quad \text{and} \quad 10^4 = 10000
\]
So,
\[
\int_{10}^{50} T^3 \, dT = \frac{6250000 - 10000}{4} = \frac{6249000}{4} = 1562250
\]
### Step 6: Substitute back to find \( Q \)
Now substituting back into the equation for \( Q \):
\[
Q = \frac{2 \cdot 1940}{281^3} \cdot 1562250
\]
Calculating \( 281^3 \):
\[
281^3 = 22196161
\]
Thus,
\[
Q = \frac{3880 \cdot 1562250}{22196161}
\]
Calculating this gives:
\[
Q \approx 273 \, \text{J}
\]
### Final Answer
The heat required to raise the temperature of 2 moles of rock salt from 10 K to 50 K is approximately \( 273 \, \text{J} \).
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