The specific heat of a metal at low temperatures varies according to `S = aT^3`, where a is a constant and T is absolute temperature. The heat energy needed to raise unit mass of the metal from temperature `T = 1 K` to `T = 2K` is

Options

A. `3a`
B. `(15a)/4`
C. `(2a)/3`
D. `(13a)/4`

Explanation

To solve the problem of finding the heat energy needed to raise the temperature of a unit mass of metal from \( T = 1 \, K \) to \( T = 2 \, K \), we can follow these steps:

### Step 1: Understand the given information
We are given that the specific heat \( S \) of the metal varies with temperature \( T \) according to the formula:
\[ S = aT^3 \]
where \( a \) is a constant.

### Step 2: Use the formula for heat transfer
The heat energy \( Q \) required to change the temperature of a substance can be expressed using the formula:
\[ Q = m \cdot S \cdot \Delta T \]
where:
- \( m \) is the mass of the substance,
- \( S \) is the specific heat,
- \( \Delta T \) is the change in temperature.

Since we are considering unit mass, we have \( m = 1 \).

### Step 3: Define the change in temperature
The change in temperature \( \Delta T \) from \( T = 1 \, K \) to \( T = 2 \, K \) is:
\[ \Delta T = T_f - T_i = 2 \, K - 1 \, K = 1 \, K \]

### Step 4: Substitute specific heat into the equation
Now, we need to express \( Q \) in terms of the specific heat. Since \( S = aT^3 \), we need to find the average specific heat over the temperature range from \( 1 \, K \) to \( 2 \, K \).

### Step 5: Calculate the average specific heat
To find the heat energy, we can integrate the specific heat over the temperature range:
\[ Q = \int_{T_i}^{T_f} S \, dT = \int_{1}^{2} aT^3 \, dT \]

### Step 6: Perform the integration
Now we perform the integration:
\[ Q = a \int_{1}^{2} T^3 \, dT \]
The integral of \( T^3 \) is:
\[ \int T^3 \, dT = \frac{T^4}{4} \]
Thus,
\[ Q = a \left[ \frac{T^4}{4} \right]_{1}^{2} \]

### Step 7: Evaluate the definite integral
Now we evaluate the definite integral:
\[ Q = a \left( \frac{2^4}{4} - \frac{1^4}{4} \right) = a \left( \frac{16}{4} - \frac{1}{4} \right) = a \left( 4 - \frac{1}{4} \right) = a \left( \frac{16 - 1}{4} \right) = a \left( \frac{15}{4} \right) \]

### Step 8: Final expression for heat energy
Thus, the heat energy needed to raise the temperature from \( 1 \, K \) to \( 2 \, K \) is:
\[ Q = \frac{15a}{4} \]