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 required to cool a 100-gram vessel of metal from 20 K to 4 K, given the specific heat capacity of the metal as a function of temperature.

### Step-by-Step Solution:

1. **Identify the mass of the metal vessel:**
- The mass \( m \) of the metal vessel is given as 100 grams.
- Convert this to kilograms:
\[ m = 100 \, \text{g} = 0.1 \, \text{kg} \]

2. **Identify the initial and final temperatures:**
- The initial temperature \( T_1 \) is 20 K.
- The final temperature \( T_2 \) is 4 K.

3. **Determine the specific heat capacity \( C_p \):**
- The specific heat capacity is given by:
\[ C_p = 32 \left( \frac{T}{400} \right)^3 \]

4. **Calculate the change in temperature \( \Delta T \):**
- The change in temperature \( \Delta T \) is:
\[ \Delta T = T_2 - T_1 = 4 \, \text{K} - 20 \, \text{K} = -16 \, \text{K} \]

5. **Calculate the work done using the first law of thermodynamics:**
- According to the first law of thermodynamics:
\[ Q = \Delta U + W \]
- Since the internal energy change \( \Delta U \) is zero for a constant temperature process, we have:
\[ Q = W \]

6. **Express the heat transfer \( Q \) in terms of specific heat capacity:**
- The heat transfer \( Q \) can be expressed as:
\[ Q = m \int_{T_1}^{T_2} C_p \, dT \]

7. **Substitute \( C_p \) into the integral:**
- Substitute \( C_p \) into the equation:
\[ Q = m \int_{20}^{4} 32 \left( \frac{T}{400} \right)^3 \, dT \]

8. **Factor out constants and set up the integral:**
- Factor out constants:
\[ Q = 32 \cdot m \cdot \frac{1}{400^3} \int_{20}^{4} T^3 \, dT \]

9. **Evaluate the integral:**
- The integral of \( T^3 \) is:
\[ \int T^3 \, dT = \frac{T^4}{4} \]
- Evaluate from 20 to 4:
\[ \left[ \frac{T^4}{4} \right]_{20}^{4} = \frac{4^4}{4} - \frac{20^4}{4} \]
- Calculate:
\[ = \frac{256}{4} - \frac{160000}{4} = 64 - 40000 = -39936 \]

10. **Calculate the total work done:**
- Substitute back into the equation for \( Q \):
\[ Q = 32 \cdot 0.1 \cdot \frac{1}{400^3} \cdot (-39936) \]
- Calculate:
\[ Q = -0.002 \, \text{kJ} \]

11. **Final answer:**
- The amount of work required to cool the vessel is:
\[ W = 0.002 \, \text{kJ} \]