Water at `50^(@)C` is filled in a closed cylindrical vessel of height `10cm` and cross sectional area `10cm^(2)` The walls of the vessel are adiabatic but the flat parts are made of `1-mm` thick aluminium `(K=200Js^(-2)m^(-1)C^(-1)` . Assume that the outside temperature is `20^(@)C` . The density of water is `1000kgm^(-3)` and the specify heat capacity of water `=4200Jkg^(-1)C^(-1)` .Estimate the time taken for the temperature to fall by `1.0^(@)C` . Make any simplifying assumption you need but specify them.

Explanation

To solve the problem of estimating the time taken for the temperature of water in a closed cylindrical vessel to fall by 1°C, we will follow these steps:

### Step 1: Gather Given Data
- Initial temperature of water, \( T_i = 50^\circ C \)
- Final temperature of water after cooling, \( T_f = 49^\circ C \)
- Height of the cylindrical vessel, \( h = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Cross-sectional area of the vessel, \( A = 10 \, \text{cm}^2 = 10 \times 10^{-4} \, \text{m}^2 \)
- Thickness of aluminum walls, \( d = 1 \, \text{mm} = 0.001 \, \text{m} \)
- Thermal conductivity of aluminum, \( K = 200 \, \text{W/m} \cdot \text{°C} \)
- Outside temperature, \( T_{outside} = 20^\circ C \)
- Density of water, \( \rho = 1000 \, \text{kg/m}^3 \)
- Specific heat capacity of water, \( c = 4200 \, \text{J/kg} \cdot \text{°C} \)

### Step 2: Calculate the Rate of Heat Transfer
The rate of heat transfer through the flat surfaces can be calculated using Fourier's law of heat conduction:

\[ Q = \frac{K \cdot A \cdot (T_i - T_{outside})}{d} \]

Substituting the values:

\[ Q = \frac{200 \, \text{W/m} \cdot \text{°C} \cdot (50 - 20) \, \text{°C} \cdot (10 \times 10^{-4} \, \text{m}^2)}{0.001 \, \text{m}} \]

Calculating this gives:

\[ Q = \frac{200 \cdot 30 \cdot 10 \times 10^{-4}}{0.001} = 6000 \, \text{W} \]

### Step 3: Calculate the Mass of Water
The mass of water in the vessel can be calculated using the formula:

\[ m = \rho \cdot V \]

Where the volume \( V \) of the cylinder is given by:

\[ V = A \cdot h = (10 \times 10^{-4} \, \text{m}^2) \cdot (0.1 \, \text{m}) = 10 \times 10^{-5} \, \text{m}^3 \]

Thus, the mass \( m \) is:

\[ m = 1000 \, \text{kg/m}^3 \cdot 10 \times 10^{-5} \, \text{m}^3 = 0.1 \, \text{kg} \]

### Step 4: Calculate the Heat Loss for a 1°C Drop
The heat loss \( \Delta Q \) required to decrease the temperature by 1°C is given by:

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

Where \( \Delta T = 1 \, \text{°C} \):

\[ \Delta Q = 0.1 \, \text{kg} \cdot 4200 \, \text{J/kg} \cdot \text{°C} \cdot 1 \, \text{°C} = 420 \, \text{J} \]

### Step 5: Calculate the Time Taken for the Temperature Drop
Using the rate of heat transfer \( Q \) and the heat loss \( \Delta Q \):

\[ \Delta Q = Q \cdot t \implies t = \frac{\Delta Q}{Q} \]

Substituting the values:

\[ t = \frac{420 \, \text{J}}{6000 \, \text{W}} = 0.07 \, \text{s} \]

### Final Answer
The estimated time taken for the temperature of the water to fall by 1°C is approximately **0.07 seconds**.

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