A flat bottomed metal tank of water is dragged along a horizontal floor at the rate of `20 ms^(-1)`. The tank is of mass 20kg and contains 1000 kg of water and all the heat produced in the dragging is conducted to the water through the bottom plate of the tank. If the bottom plate has an effective area of conduction 1 m^2 and a thickness 5 cm and the temperature of the water in the tank remains constant at `50^(@)C`, calculate the temperature of the bottom surface of the tank, given the coefficient of friction between the tank and the floor is 0.343 and K for the material of the tank is `25 cal m^(-1) s^(-1) K^(-1).`

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the total force due to friction
The force of friction (F) can be calculated using the formula:
\[ F = \mu \cdot M \cdot g \]
where:
- \( \mu = 0.343 \) (coefficient of friction)
- \( M = 20 \, \text{kg} + 1000 \, \text{kg} = 1020 \, \text{kg} \) (total mass of the tank and water)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)

Calculating the force:
\[ F = 0.343 \cdot 1020 \cdot 10 = 3499.4 \, \text{N} \]

### Step 2: Calculate the power generated due to friction
Power (P) can be calculated using the formula:
\[ P = F \cdot v \]
where:
- \( v = 20 \, \text{m/s} \) (velocity)

Calculating the power:
\[ P = 3499.4 \cdot 20 = 69988 \, \text{W} \]

### Step 3: Convert power to heat energy per second
Since power is already in watts (joules per second), we can directly use this value as the heat energy produced per second:
\[ H = 69988 \, \text{J/s} \]

### Step 4: Apply the heat conduction formula
Using the heat conduction formula:
\[ Q = \frac{k \cdot A \cdot (T_1 - T_2)}{d} \]
where:
- \( Q = H = 69988 \, \text{J/s} \)
- \( k = 25 \, \text{cal/m/s/K} = 25 \cdot 4.2 \, \text{J/m/s/K} = 105 \, \text{J/m/s/K} \) (convert calories to joules)
- \( A = 1 \, \text{m}^2 \) (area)
- \( d = 5 \, \text{cm} = 0.05 \, \text{m} \) (thickness)
- \( T_1 = 50 \, \text{°C} \) (temperature of water)
- \( T_2 \) is the temperature of the bottom surface of the tank.

Rearranging the formula to solve for \( T_2 \):
\[ T_1 - T_2 = \frac{Q \cdot d}{k \cdot A} \]

### Step 5: Substitute values and solve for \( T_2 \)
Substituting the known values:
\[ 50 - T_2 = \frac{69988 \cdot 0.05}{105 \cdot 1} \]
\[ 50 - T_2 = \frac{3499.4}{105} \]
\[ 50 - T_2 = 33.33 \]

Now, solving for \( T_2 \):
\[ T_2 = 50 - 33.33 = 16.67 \, \text{°C} \]

### Final Answer
The temperature of the bottom surface of the tank is approximately:
\[ T_2 \approx 16.67 \, \text{°C} \]

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