water in a container having 2 mm thick walls having thermal conductivity 0.5.container is kept in a melting ice bath at 0C.total surface area in contact with water is 0.05 m^2.a wheel is clamped inside the water and is coupled toa block of mass M.as the block goes down,the wheel rotates.it is found that after sumtym a steady state is reached in which the block goes down witha constant speed of 10cm/s and the temperature of the water remains constant at 1 C .Find the mass M of the block.assume heat flows out of the water only through the walls in contact

To determine the mass \( M \) of the block in this scenario, we need to analyze the heat transfer occurring through the container's walls and the energy balance involved in the system. Since the water temperature remains constant at 1°C, we can infer that the heat lost by the water is equal to the work done by the block as it descends. Let's break this down step by step.

Understanding Heat Transfer

The heat transfer through the container walls can be calculated using Fourier's law of heat conduction. The formula for heat transfer \( Q \) through a material is given by:

Q = k \cdot A \cdot \frac{\Delta T}{d}

• k = thermal conductivity of the material (0.5 W/m·K)
• A = surface area in contact with water (0.05 m²)
• \(\Delta T\) = temperature difference between the water and the melting ice bath (1°C - 0°C = 1°C)
• d = thickness of the walls (0.002 m)

Calculating Heat Transfer Rate

Now, substituting the values into the formula:

Q = 0.5 \cdot 0.05 \cdot \frac{1}{0.002}

Calculating this gives:

Q = 0.5 \cdot 0.05 \cdot 500 = 12.5 W

This means that the heat is being lost from the water at a rate of 12.5 watts.

Relating Heat Loss to Work Done

Since the block descends at a constant speed of 10 cm/s, the work done by the block as it descends can be expressed as:

W = M \cdot g \cdot h

Where:

• M = mass of the block
• g = acceleration due to gravity (approximately 9.81 m/s²)
• h = height descended in a given time

Since the block moves down at a constant speed, we can relate the height descended over time \( t \) to the speed:

h = v \cdot t

Substituting \( v = 0.1 \) m/s (10 cm/s) gives:

h = 0.1 \cdot t

Equating Heat Loss to Work Done

At steady state, the power lost by the water equals the power used to do work on the block:

Q = M \cdot g \cdot \frac{h}{t}

Substituting \( h = 0.1 \cdot t \) into the equation gives:

12.5 = M \cdot 9.81 \cdot 0.1

Now, solving for \( M \):

M = \frac{12.5}{9.81 \cdot 0.1} = \frac{12.5}{0.981} \approx 12.74 \text{ kg}

Final Result

The mass \( M \) of the block is approximately 12.74 kg. This calculation shows how the principles of heat transfer and mechanical work are interconnected in this system, illustrating the balance between energy lost and energy used effectively.