Question

The molar heat capacity of rock salt at low temperature varies with temperature according to Debye's \( T^3 \) law. Thus \( C = k \frac{T^3}{\theta^3} \) where \( k = 1940 \, \text{Jmol}^{-1} \text{K}^{-1} \) and \( \theta = 281 \, \text{K} \). Calculate how much heat is required to raise the temperature of 2 moles of rock salt from 10 K to 50 K:

• A. 800 J
• B. 373 J
• C. 273 J
• D. None of these

Hint:

To calculate heat using Debye's law, use the formula for molar heat capacity and integrate over the temperature range.

Answer 1

The Correct Option is C

Step 1: Write the expression for heat required.
The heat required \( Q \) is given by: \[ Q = n \int_{T_1}^{T_2} C(T) \, dT \] where \( n = 2 \, \text{moles} \), and \( C(T) = k \frac{T^3}{\theta^3} \) is the molar heat capacity.

Step 2: Set up the integral.
Substitute the expression for \( C(T) \) into the integral: \[ Q = n \int_{T_1}^{T_2} k \frac{T^3}{\theta^3} \, dT \] \[ Q = 2 \times 1940 \times \frac{1}{281^3} \int_{10}^{50} T^3 \, dT \]

Step 3: Perform the integration.
Integrating \( T^3 \) from 10 K to 50 K: \[ \int T^3 \, dT = \frac{T^4}{4} \] Substitute the limits: \[ Q = 2 \times 1940 \times \frac{1}{281^3} \left( \frac{50^4}{4} - \frac{10^4}{4} \right) \]

Step 4: Final calculation.
The final heat required is approximately: \[ Q = 273 \, \text{J} \]