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

Explanation

To solve the problem, we need to calculate the work required to cool a 100-gram metal vessel from 20 K to 4 K using the specific heat capacity formula provided. Here’s a step-by-step breakdown of the solution:

### Step 1: Convert Mass to Kilograms
The mass of the vessel is given as 100 grams. We need to convert this to kilograms for consistency with the specific heat capacity units.

\[ m = 100 \text{ grams} = 0.1 \text{ kg} \]

**Hint:** Remember to convert grams to kilograms by dividing by 1000.

### Step 2: Write Down the Specific Heat Capacity Formula
The specific heat capacity \( C_p \) of the metal at low temperature \( T \) is given by:

\[ C_p = 32 \left( \frac{T}{400} \right)^3 \text{ kJ/(K kg)} \]

**Hint:** Ensure you understand how to substitute the temperature \( T \) into the formula.

### Step 3: Set Up the Heat Transfer Equation
The heat \( dq \) absorbed or released during a temperature change can be expressed as:

\[ dq = m C_p dT \]

Substituting for \( C_p \):

\[ dq = 0.1 \times 32 \left( \frac{T}{400} \right)^3 dT \]

**Hint:** Remember that \( dT \) is the change in temperature, which will be integrated over the temperature range.

### Step 4: Define the Limits of Integration
We need to cool the vessel from 20 K to 4 K. Thus, the limits for integration will be from \( T = 20 \) K to \( T = 4 \) K.

**Hint:** Be careful with the limits; the upper limit is the initial temperature and the lower limit is the final temperature.

### Step 5: Integrate the Heat Transfer Equation
Now we integrate \( dq \) from 20 K to 4 K:

\[ W = \int_{20}^{4} 0.1 \times 32 \left( \frac{T}{400} \right)^3 dT \]

This simplifies to:

\[ W = 0.1 \times 32 \times \frac{1}{400^3} \int_{20}^{4} T^3 dT \]

**Hint:** The integral of \( T^3 \) is \( \frac{T^4}{4} \).

### Step 6: Calculate the Integral
Calculating the integral:

\[ \int T^3 dT = \frac{T^4}{4} \bigg|_{20}^{4} = \frac{4^4}{4} - \frac{20^4}{4} \]

Calculating the values:

\[ = \frac{256}{4} - \frac{160000}{4} = 64 - 40000 = -39936 \]

**Hint:** Ensure you perform the calculations carefully, especially with powers.

### Step 7: Substitute Back into the Work Equation
Now substitute back into the work equation:

\[ W = 0.1 \times 32 \times \frac{1}{400^3} \times (-39936) \]

Calculating \( W \):

\[ W = 0.1 \times 32 \times \frac{-39936}{64000000} \]

### Step 8: Final Calculation
Calculating the final value:

\[ W = \frac{-127.488}{640000} = -0.0001992 \text{ kJ} \approx 0.002 \text{ kJ} \]

Since work done is typically expressed as a positive quantity, we take the absolute value.

### Conclusion
The amount of work required to cool the vessel is:

\[ W \approx 0.002 \text{ kJ} \]

**Final Answer:** \( 0.002 \text{ kJ} \)

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