The change in melting temperature of ice (in K ) when it is subjected to a pressure of 100 atm is _ Given: density of ice is , density of water is and latent heat of fusion is .
Ground off to two decimal places).

Ask by Daniel Craig. in India

Jan 21,2025

Question

The change in melting temperature of ice (in K ) when it is subjected to a pressure of 100 atm is _ Given: density of ice is , density of water is and latent heat of fusion is .
Ground off to two decimal places).

Ask by Daniel Craig. in India

Jan 21,2025

UpStudy AI · Accepted Answer

To determine the change in the melting temperature of ice when subjected to a pressure of 100 atmospheres (atm), we can use the **Clausius-Clapeyron equation**. This equation relates the change in temperature to the change in pressure for a phase transition, such as melting.

### Given:
- **Density of ice ($$\rho_{\text{ice}}$$)** = 0.92 g/cm³
- **Density of water ($$\rho_{\text{water}}$$)** = 1 g/cm³
- **Latent heat of fusion ($$L$$)** = 80 cal/g
- **Pressure change ($$\Delta P$$)** = 100 atm
- **Standard melting temperature of ice ($$T$$)** = 273.15 K

### Steps:

1. **Calculate the Change in Volume ($$\Delta V$$)**:
   $$
   \Delta V = \frac{1}{\rho_{\text{water}}} - \frac{1}{\rho_{\text{ice}}} = \frac{1}{1} - \frac{1}{0.92} = -0.08696 \, \text{cm}^3/\text{g}
   $$
   Convert to m³/g:
   $$
   \Delta V = -0.08696 \times 10^{-6} \, \text{m}^3/\text{g}
   $$

2. **Convert Latent Heat to Consistent Units**:
   $$
   L = 80 \, \text{cal/g} \times 4.184 \, \text{J/cal} = 334.72 \, \text{J/g}
   $$

3. **Apply the Clausius-Clapeyron Equation**:
   $$
   \frac{dT}{dP} = \frac{T \Delta V}{L} = \frac{273.15 \times (-0.08696 \times 10^{-6})}{334.72} \approx -7.18 \times 10^{-3} \, \text{K/atm}
   $$

4. **Calculate the Total Change in Temperature ($$\Delta T$$)**:
   $$
   \Delta T = \frac{dT}{dP} \times \Delta P = -7.18 \times 10^{-3} \times 100 = -0.72 \, \text{K}
   $$

### Conclusion:
The melting temperature of ice decreases by **0.72 K** when subjected to a pressure of 100 atm.

**Answer:**  
The melting temperature decreases by 0.72 K under 100 atm of pressure.
The melting temperature of ice decreases by 0.72 K when subjected to a pressure of 100 atm.

The change in melting temperature of ice (in K ) when it is subjected to a pressure of 100 atm is _ Given: density of ice is , density of water is and latent heat of fusion is .
Ground off to two decimal places).

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The change in melting temperature of ice (in K ) when it is subjected to a pressure of 100 atm is _ Given: density of ice is , density of water is and latent heat of fusion is . Ground off to two decimal places).

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The change in melting temperature of ice (in K ) when it is subjected to a pressure of 100 atm is _ Given: density of ice is , density of water is and latent heat of fusion is .
Ground off to two decimal places).