Knowee
Questions
Features
Study Tools

he densities of ice and liquid water at 1 atm (101,325 N/m2) and 0°C (273 K) are 0.917 g/c.c. (0.917 x IO3 kg/m3) and 0.9998 g/c.c. (0.9998 xio3 respectively. The heat of fusion of water is 80 cal/g (334.72 x 10 J/kg). o Calculate the melting point of ice at 0.5 atm (50,662.5 N/m ) and 101 atm (102,338 X IO2 N/m2).

Question

he densities of ice and liquid water at 1 atm (101,325 N/m2) and 0°C (273 K) are 0.917 g/c.c. (0.917 x IO3 kg/m3) and 0.9998 g/c.c. (0.9998 xio3 respectively. The heat of fusion of water is 80 cal/g (334.72 x 10 J/kg). o Calculate the melting point of ice at 0.5 atm (50,662.5 N/m ) and 101 atm (102,338 X IO2 N/m2).

🧐 Not the exact question you are looking for?Go ask a question

Solution

The melting point of a substance is influenced by pressure. The Clausius-Clapeyron equation is used to calculate the change in melting point with pressure. The equation is:

ΔT/ΔP = -TΔv/ΔH

where: ΔT is the change in temperature, ΔP is the change in pressure, T is the absolute temperature, Δv is the change in specific volume (v_f - v_i, where v_f is the specific volume of the fluid and v_i is the specific volume of the ice), and ΔH is the enthalpy change of fusion.

Given: T = 273 K, ΔH = 334.72 x 10^3 J/kg, v_f = 1/0.9998 x 10^3 kg/m^3, v_i = 1/0.917 x 10^3 kg/m^3.

We can calculate Δv = v_f - v_i.

Then, we can use the Clausius-Clapeyron equation to find ΔT for the change in pressure from 1 atm to 0.5 atm and from 1 atm to 101 atm.

Please note that this calculation assumes that ΔH, v_f, and v_i do not change with temperature or pressure, which is an approximation. The actual change in melting point may be slightly different.

This problem has been solved

Similar Questions

18 gm of ice is converted into water at 0°C and 1 atm. The entropies of  and  are 38.2 and 60 J/mol K respectively. The enthalpy change for this conversion is :

Mercury melts at -38.87°C (234.13 K) at 1 atm (101,325 N/m2) pressure. The densities of liquid and solid mercury are 13.69 g/c.c. (13.69 x IO3 kg/m3) 33 and 14.19 g/c.c. (14.19 x 10 kg/m ) respectively at the normal melting point. 7 The heat of fusion of mercury is 2.33 cal/g (9.75 x 10 J/kg). Calcu

The densities of liquid and solid bismuth are 10.0 g/c.c. (10x10 kg/m ) and 9.673 g/c.c. (9.673 x IO3 kg/m3) respectively at the normal melting point 270°C (543 K). The heat of fusion is 2.633 kcal/mole (11.02 kJ/mol). Calculate the melting point of bismuth under a pressure of 100 atm (101,325 x 22 10 N/m ) . Atomic weight of bismuth is 209.

The specific heat of ice is 2.10 kJ/kg °C, the heat of fusion for ice at 0 °C is 333.7 kJ/kg, the specific heat of water 4.186 kJ/kg °C, and the heat of vaporization of water at 100 °C is 2,256 kJ/kg. What is the final equilibrium temperature when 5.00 grams of ice at −15.0 °C is mixed with 40.0 grams of water at 75.0 °C?

50 g of ice at 0°C is added to 300 g of a liquid at 30°C. What will be the finaltemperature of the mixture when all the ice has melted? The specific heat capacity ofthe liquid is 265Jg-1°C-1 while that of water is 4 2 Jg-1°C-1.Specific latent heat of fusion of ice =336Jg-1 .[2]

1/3

Upgrade your grade with Knowee

Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.