The specific heat capacity of a metal at low temperature (T) is given as
`C_(p)(kJK^(-1) kg^(-1)) =32((T)/(400))^(3)`
A 100 gram vessel of this metal is to be cooled from `20^(@)K` to `4^(@)K` by a special refrigerator operating at room temperaturte `(27^(@)C)` . The amount of work required to cool the vessel is

Options

A. equal to `0.002` kJ
B. greater than `0.148` kJ
C. between `0.148` kJ and `0.028` kJ
D. less than `0.028` kJ

Explanation

To solve the problem, we need to calculate the work done to cool a 100-gram vessel of metal from 20 K to 4 K, given the specific heat capacity formula. Let's break this down step by step.

### Step 1: Understand the given data
- Mass of the metal vessel, \( m = 100 \, \text{grams} = 0.1 \, \text{kg} \)
- Initial temperature, \( T_i = 20 \, \text{K} \)
- Final temperature, \( T_f = 4 \, \text{K} \)
- Specific heat capacity, \( C_p = 32 \left( \frac{T}{400} \right)^3 \, \text{kJ/(K kg)} \)

### Step 2: Set up the equation for work done
The work done \( W \) to cool the vessel can be expressed as the heat removed from the vessel:
\[ W = -\Delta Q = -\int_{T_i}^{T_f} m C_p \, dT \]
Where \( \Delta Q \) is the heat removed.

### Step 3: Substitute \( C_p \) into the equation
Substituting the expression for \( C_p \):
\[ W = -\int_{20}^{4} 0.1 \cdot 32 \left( \frac{T}{400} \right)^3 \, dT \]

### Step 4: Simplify the integral
We can factor out constants from the integral:
\[ W = -0.1 \cdot 32 \cdot \frac{1}{400^3} \int_{20}^{4} T^3 \, dT \]

### Step 5: Calculate the integral
The integral of \( T^3 \) is:
\[ \int T^3 \, dT = \frac{T^4}{4} \]
Now, we evaluate the definite integral:
\[ \int_{20}^{4} T^3 \, dT = \left[ \frac{T^4}{4} \right]_{20}^{4} = \frac{4^4}{4} - \frac{20^4}{4} \]
Calculating the values:
\[ 4^4 = 256 \quad \text{and} \quad 20^4 = 160000 \]
Thus,
\[ \int_{20}^{4} T^3 \, dT = \frac{256}{4} - \frac{160000}{4} = 64 - 40000 = -39936 \]

### Step 6: Substitute back into the work equation
Now substituting back into the equation for \( W \):
\[ W = -0.1 \cdot 32 \cdot \frac{1}{400^3} \cdot (-39936) \]
This simplifies to:
\[ W = 0.1 \cdot 32 \cdot \frac{39936}{400^3} \]

### Step 7: Calculate the numerical value
Calculating \( 400^3 = 64000000 \):
\[ W = 0.1 \cdot 32 \cdot \frac{39936}{64000000} \]
Calculating \( 32 \cdot 39936 = 1279872 \):
\[ W = 0.1 \cdot \frac{1279872}{64000000} = \frac{127987.2}{640000} = 0.002 \, \text{kJ} \]

### Final Answer
The amount of work required to cool the vessel is:
\[ \boxed{0.002 \, \text{kJ}} \]