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Question

Two blocks of ice when pressed together join to form one block because

The correct answer is Melting point of ice decreases with increase of pressure

Understanding Why Ice Blocks Join Under Pressure

When two blocks of ice are pressed together, they often stick or join to form a single block. This interesting phenomenon is related to how pressure affects the melting point of ice.

The Effect of Pressure on Melting Point

For most substances, an increase in pressure raises the melting point. However, water (and therefore ice) is an exception to this rule. For ice, increasing the pressure actually lowers its melting point.

This unusual behavior is due to the fact that water is denser than ice. When ice melts under pressure, the resulting water occupies less volume. According to Le Chatelier's principle, a system at equilibrium will shift in a direction that relieves the applied stress. Applying pressure is a stress that favors the phase (liquid) which occupies less volume. Thus, increasing pressure on ice favors melting, which means the melting temperature must decrease.

The relationship between pressure and melting temperature is described by the Clausius-Clapeyron equation, which qualitatively shows that for substances like water where the solid is less dense than the liquid, the melting temperature decreases with increasing pressure.

$$\frac{dP}{dT} = \frac{L}{T\Delta V}$$

Here, $P$ is pressure, $T$ is temperature, $L$ is the latent heat of fusion, and $\Delta V$ is the change in volume during melting. For water, $\Delta V = V_{liquid} - V_{solid} < 0$, so $\frac{dP}{dT} < 0$. This negative slope indicates that increasing pressure (positive $dP$) leads to a decrease in melting temperature (negative $dT$).

Regelation: The Process of Joining Ice Blocks

The joining of ice blocks under pressure is a demonstration of a phenomenon called regelation. Here's how it works:

  • When you press two ice blocks together, you apply pressure to the points of contact.
  • This applied pressure lowers the melting point of the ice at these points.
  • If the temperature of the ice is at or slightly below 0°C, the lowered melting point at the contact surfaces falls below the actual temperature of the ice.
  • This causes a thin layer of ice at the contact points to melt into water, even though the bulk of the ice remains solid.
  • When the pressure is removed (or reduced as the contact area increases), the melting point at these surfaces returns to its normal value (approximately 0°C at standard atmospheric pressure).
  • Since the surrounding temperature is still at or below 0°C, the thin layer of water refreezes, solidifying and joining the two blocks into one.

Evaluating the Options

  • Option 1: of heat produced during pressing - While some minor friction might produce a tiny amount of heat, this is not the primary reason for the blocks joining. The core mechanism involves a phase change induced by pressure, not temperature increase.
  • Option 2: of cold produced during pressing - Pressing does not produce cold. This option is incorrect.
  • Option 3: Melting point of ice decreases with increase of pressure - This correctly describes the effect of pressure on ice's melting point, which is the fundamental principle behind regelation and the joining of ice blocks.
  • Option 4: Melting point of ice increases with increase of pressure - This is incorrect for ice. For most substances, the melting point increases with pressure, but water is an exception.

Therefore, the reason two blocks of ice join when pressed together is because the melting point of ice decreases with an increase in pressure, leading to melting and subsequent refreezing.

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Important Questions from Thermodynamics

  1. If the work done on the system or by the system· is zero, which one of the following statements for a gas kept at a certain volume is correct?

  2. A system that does NOT allow exchange of heat with its surrounding is called

  3. A system that does NOT allow exchange of heat with its surrounding is called

  4. For a certain reaction, ΔG θ = -45 kJ/mol and ΔH θ = -90 kJ/mol at 0 °C. What is the minimum temperature at which the reaction will become spontaneous, assuming that ΔH θ  and ΔS θ  are independent of temperature?

  5. Which of the following statements correctly describes the thermodynamic classification of entropy?
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