Concepts Of Physics MCQ Edition [Volume 2]
- Exam
- Concepts Of Physics MCQ Edition [Volume 2]
- Subject
- Physics
- Chapter
- Heat Transfer
Concepts Of Physics MCQ Edition [Volume 2] · Physics · Heat Transfer
Water is held in a container with 2.0 -mm thick walls composed of a...
Water is held in a container with -mm thick walls composed of a material having a thermal conductivity of 0.50 \text{ W m}^{-1} ^{}^{{}^{\circ}}\textC^{-1}, as shown in the figure. The container is immersed in a melting-ice bath at 0^{}^{{}^{\circ}}\text{C}. The total surface area of the walls in contact with the water is . A wheel is clamped inside the water and is coupled to a block of mass . The wheel rotates as the block descends. After some time, a steady state is established where the block descends with a constant speed of and the water temperature remains steady at 1.0^{}^{{}^{\circ}}\text{C}. Determine the mass of the block, assuming heat flows out of the water only through the contact walls. Use .
Options
- A
- B
- C
- D
Correct answer
B.
Step-by-step solution
At steady state, the rate of mechanical work done by the descending block is equal to the rate of heat conducted through the walls of the container.
The mechanical power generated by the block is:
The rate of heat conduction through the container walls is given by Fourier's law:
Equating the mechanical power to the rate of heat conduction:
Mgv = \dfrac{kA(T_w - T_0)}{d}k = 0.50 \text{ W m}^{-1} ^{}^{{}^{\circ}}\textC^{-1}A = 0.05 \text{ m}^2T_w = 1.0^{}^{{}^{\circ}}\text{C}T_0 = 0^{}^{{}^{\circ}}\text{C}d = 2.0 \text{ mm} = 2.0 \times 10^{-3} \text{ m}v = 10 \text{ cm s}^{-1} = 0.1 \text{ m s}^{-1}g = 10 \text{ m s}^{-2}M \times 10 \times 0.1 = \dfrac{0.50 \times 0.05 \times (1.0 - 0)}{2.0 \times 10^{-3}}M \times 1 = \dfrac{0.025}{0.002}M = 12.5 \text{ kg}
The mechanical power generated by the block is:
The rate of heat conduction through the container walls is given by Fourier's law:
Equating the mechanical power to the rate of heat conduction:
Mgv = \dfrac{kA(T_w - T_0)}{d}k = 0.50 \text{ W m}^{-1} ^{}^{{}^{\circ}}\textC^{-1}A = 0.05 \text{ m}^2T_w = 1.0^{}^{{}^{\circ}}\text{C}T_0 = 0^{}^{{}^{\circ}}\text{C}d = 2.0 \text{ mm} = 2.0 \times 10^{-3} \text{ m}v = 10 \text{ cm s}^{-1} = 0.1 \text{ m s}^{-1}g = 10 \text{ m s}^{-2}M \times 10 \times 0.1 = \dfrac{0.50 \times 0.05 \times (1.0 - 0)}{2.0 \times 10^{-3}}M \times 1 = \dfrac{0.025}{0.002}M = 12.5 \text{ kg}