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Grade upto college level Thermal Physics

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

Profile image of Deepak Patra
12 Years agoGrade upto college level
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ApprovedApproved Tutor Answer0 Years ago

To solve this problem, we need to analyze the heat transfer occurring through the walls of the container and how it relates to the motion of the block and the wheel inside the water. The key points to consider are the thermal conductivity of the container, the temperature difference, and the steady-state conditions of the system.

Understanding Heat Transfer

Heat transfer through the walls of the container can be described using Fourier's law of heat conduction, which states that the rate of heat transfer (Q) through a material is proportional to the temperature difference (ΔT) across the material and the surface area (A) through which the heat is being transferred. The formula can be expressed as:

Q = k * A * (ΔT / d)

Where:

  • Q = rate of heat transfer (in watts)
  • k = thermal conductivity of the material (in W/m·K)
  • A = surface area (in m²)
  • ΔT = temperature difference (in °C or K)
  • d = thickness of the material (in meters)

Given Values

From the problem, we have:

  • Thermal conductivity, k = 0.5 W/m·K
  • Wall thickness, d = 2 mm = 0.002 m
  • Surface area in contact with water, A = 0.05 m²
  • Temperature of the ice bath, T_ice = 0°C
  • Temperature of water, T_water = 1°C

Calculating the Heat Transfer Rate

The temperature difference between the water and the ice bath is:

ΔT = T_water - T_ice = 1°C - 0°C = 1°C

Now, substituting the values into the heat transfer equation:

Q = 0.5 W/m·K * 0.05 m² * (1 K / 0.002 m)

Q = 0.5 * 0.05 * 500 = 12.5 W

Relating Heat Transfer to the Block's Motion

In steady state, the heat lost by the water (12.5 W) is equal to the work done by the block as it descends. The block moves down at a constant speed of 10 cm/s, which means the force exerted by the block (due to its weight) is balanced by the torque produced by the wheel's rotation.

The power (P) exerted by the block as it descends can be expressed as:

P = F * v

Where:

  • P = power (in watts)
  • F = force (in newtons)
  • v = velocity (in m/s)

Given that the velocity v = 10 cm/s = 0.1 m/s, we can express the force as:

F = M * g

Where g is the acceleration due to gravity (approximately 9.81 m/s²). Thus, we can rewrite the power equation as:

P = M * g * v

Setting Up the Equation

Since we know the power exerted by the block must equal the heat transfer rate:

12.5 W = M * 9.81 m/s² * 0.1 m/s

Now, we can solve for the mass M:

M = 12.5 W / (9.81 m/s² * 0.1 m/s)

M = 12.5 / 0.981 = 12.74 kg

Final Result

The mass of the block M is approximately 12.74 kg. This means that for the system to maintain a constant temperature of 1°C while the block descends at a steady speed, the block must have this mass. This balance between heat loss and mechanical work illustrates the principles of thermodynamics and energy conservation in a practical scenario.