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One end of a metal rod is dipped in boil...

One end of a metal rod is dipped in boiling water and the other is dipped in melting ice.

A

all parts of the rod are in thermal equilibrium with each other

B

we can assign a temperature to the rod

C

we can assign temperature to the rod after steady state is reached.

D

the state of the rod does not change after steady state is reached

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the situation of a metal rod with one end in boiling water and the other end in melting ice. Here’s a step-by-step breakdown of the solution: ### Step 1: Understanding the Setup - One end of the metal rod is in boiling water (at 100°C). - The other end is in melting ice (at 0°C). - The rod is made of a material that conducts heat. **Hint:** Think about the temperature differences and how they will affect the rod. ### Step 2: Heat Transfer Mechanism - Heat will transfer from the hot end (boiling water) to the cold end (melting ice) through conduction. - Conduction is the process where heat energy is transferred through collisions between particles in a material. **Hint:** Recall the definition of conduction and how it occurs in solids. ### Step 3: Establishing Steady State - As heat flows from the hot end to the cold end, the temperature along the rod will change. - Eventually, the rod will reach a steady state where the temperature distribution becomes constant over time, but not uniform. **Hint:** What does "steady state" mean in terms of temperature changes? ### Step 4: Temperature Distribution - In steady state, the temperature will not be the same throughout the rod. - The temperature will be highest at the end in boiling water and lowest at the end in melting ice. - The temperature will vary linearly between these two extremes. **Hint:** Consider how temperature gradients work in a conductive material. ### Step 5: Conclusion on Thermal Equilibrium - Once steady state is reached, the temperature distribution remains constant over time. - However, the system is not in thermal equilibrium in the sense that all parts of the rod have the same temperature; rather, they have a stable temperature gradient. **Hint:** What does thermal equilibrium imply about the temperatures of different parts of the rod? ### Final Answer The state of the rod does not change after steady state is reached, but the temperature is different at various points along the rod. ### Summary of Key Points - Heat transfer occurs via conduction from the hot end to the cold end. - Steady state is reached when the temperature distribution is constant over time. - The rod has a non-uniform temperature distribution, with the highest temperature at the boiling water end and the lowest at the melting ice end.

To solve the problem, we need to analyze the situation of a metal rod with one end in boiling water and the other end in melting ice. Here’s a step-by-step breakdown of the solution: ### Step 1: Understanding the Setup - One end of the metal rod is in boiling water (at 100°C). - The other end is in melting ice (at 0°C). - The rod is made of a material that conducts heat. **Hint:** Think about the temperature differences and how they will affect the rod. ...
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Which melts in boiling water :

A metallic rod of cross-sectional area 20 cm^(2) , with the lateral surface insulated to prevent heat loss, has one end immersed in boiling water and the other in ice water mixture. The heat conducted through the rod melts the ice at the rate of 1 gm for every 84 sec. The thermal conductivity of the rod is 160 Wm^(-1)K^(-1) . Latent heat of ice=80 cal/gm, 1 ca=4.2 joule. What is the length (in m) of the rod?

Knowledge Check

  • One end of a 0.25 m long metal bar is in steam and the other is in contact with ice . If 12 g of ice melts per minute, what is the thermal conductivity of the metal? Given cross-section of the bar =5xx10^(-4) m^(2) and latent heat of ice is 80 cal g^(-1)

    A
    `20 cal s^(-1) m^(-1).^(@)C^(-1)`
    B
    `10 cal s^(-1) m^(-1) .^(@)C^(-1)`
    C
    `40 cal s^(-1) m^(-1) .^(@)C^(-1)`
    D
    `80 cal s^(-1) m^(-1) .^(@)C^(-1)`
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