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 find the heat energy needed to raise the unit mass of the metal from temperature \( T = 1 \, K \) to \( T = 2 \, K \), we will use the formula for heat energy, which is given by:

\[ Q = m \cdot S \cdot \Delta T \]

where:
- \( Q \) is the heat energy,
- \( m \) is the mass of the substance,
- \( S \) is the specific heat capacity,
- \( \Delta T \) is the change in temperature.

### Step 1: Identify the parameters
We know that:
- The specific heat \( S = aT^3 \),
- The mass \( m = 1 \, kg \) (unit mass),
- The initial temperature \( T_1 = 1 \, K \),
- The final temperature \( T_2 = 2 \, K \).

### Step 2: Calculate the change in temperature
The change in temperature \( \Delta T \) is given by:

\[ \Delta T = T_2 - T_1 = 2 \, K - 1 \, K = 1 \, K \]

### Step 3: Set up the integral for heat energy
Since the specific heat \( S \) varies with temperature, we need to integrate to find the total heat energy required to raise the temperature from \( T_1 \) to \( T_2 \):

\[ Q = \int_{T_1}^{T_2} S \, dT = \int_{1}^{2} aT^3 \, dT \]

### Step 4: Perform the integration
We can factor out the constant \( a \):

\[ Q = a \int_{1}^{2} T^3 \, dT \]

Now, we integrate \( T^3 \):

\[ \int T^3 \, dT = \frac{T^4}{4} \]

So, we evaluate the definite integral:

\[ Q = a \left[ \frac{T^4}{4} \right]_{1}^{2} = a \left( \frac{2^4}{4} - \frac{1^4}{4} \right) \]

Calculating the values:

\[ Q = a \left( \frac{16}{4} - \frac{1}{4} \right) = a \left( 4 - \frac{1}{4} \right) = a \left( \frac{16 - 1}{4} \right) = a \left( \frac{15}{4} \right) \]

### Step 5: Final expression for heat energy
Thus, the heat energy needed to raise the unit mass of the metal from \( T = 1 \, K \) to \( T = 2 \, K \) is:

\[ Q = \frac{15a}{4} \]

### Conclusion
The heat energy needed is \( \frac{15a}{4} \).

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