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 \( 0.92 \mathrm{~g} / \mathrm{cm}^{3} \), density of water is \( 1 \mathrm{~g} / \mathrm{cm}^{3} \) and latent heat of fusion is \( 80 \mathrm{cal} / \mathrm{g} \). Ground off to two decimal places).

To find the change in melting temperature of ice under pressure, we can use the Clausius-Clapeyron equation, which relates changes in pressure to changes in temperature during phase changes. The formula is given by: \[ \frac{dT}{dP} = \frac{L}{T \Delta V} \] where \( L \) is the latent heat of fusion, \( T \) is the melting temperature of ice (approximately \( 273 \, K \) at standard conditions), and \( \Delta V \) is the change in volume. 1. First, let's calculate \( \Delta V \): The volume of ice can be calculated as: \[ V_{ice} = \frac{m}{\rho_{ice}} = \frac{1 \, \text{g}}{0.92 \, \text{g/cm}^3} \approx 1.087 \, \text{cm}^3 \] The volume of water: \[ V_{water} = \frac{m}{\rho_{water}} = \frac{1 \, \text{g}}{1 \, \text{g/cm}^3} = 1 \, \text{cm}^3 \] The change in volume (\( \Delta V \)) during melting is: \[ \Delta V = V_{water} - V_{ice} = 1 - 1.087 \approx -0.087 \, \text{cm}^3 \] 2. Now substituting into the Clausius-Clapeyron equation: Take \( L = 80 \, \text{cal/g} = 80 \times 4.184 \, \text{J/g} \approx 334.72 \, \text{J/g} \) (since \( 1 \, \text{cal} = 4.184 \, \text{J} \)). The change in temperature for a pressure change of \( \Delta P = 100 \, \text{atm} \): \[ \frac{dT}{dP} = \frac{334.72}{273 \times (-0.087 \times 10^{-6})} \quad \text{(convert cm}^3\text{ to m}^3\text{)} \] \[ dT = \frac{334.72 \times 100 \times 101325}{273 \times (-0.087 \times 10^{-6})} \approx -0.573 \, K \] Thus, the change in melting temperature of ice when it is subjected to a pressure of 100 atm is approximately \(-0.57 \, K\). Grounded to two decimal places, the answer is \(-0.57\).