PLEASE ANSWER EVERYTHING THROUGHLY Relevant data: At T = 0 degree C and P= 1 atm

Question

PLEASE ANSWER EVERYTHING THROUGHLY

Relevant data: At T = 0 degree C and P= 1 atm, the density of ice is 917 kg/m^3 and that of water is 1000 kg/m^3 The latent heat (required to melt ice to water) is 333 kJ/kg. Use the Clausius-Clapeyron equation to determine the slope of the phase boundary between water and ice at the temperature and pressure quoted above. Express your answer in units bar/K, where 1 bar =10^5 N/m^2. How much pressure would you have to put on an ice cube to make it melt at -1 degree C? Approximately how deep under the surface of a glacier would you have to be before the weight of the ice above gives the pressure you found in part (a)? (Answer: 1:5 km) This question is interesting because some authors have claimed that the high pressures at the bottom of glaciers melts the ice, and the resulting liquid water helps to lubricate the movement of the glacier. How realistic do you think this explanation is?

Explanation / Answer

(a) The Clausius-Clapeyron relation applies to the slope of any phase boundary line on a Phase Transfer diagram:

dP/dT = L/T(1/l- 1/s)

dP/dT = 333 kJ/kg/[273 K (1/1000-1/917) m3/kg] = -13476.37 kJ K-1 m-3 = -134.76 * 10^5 N m-2 K-1 = -134.76 bar/K

Infinitesimal changes in pressure and temperature

dP/dT = L/T(1/l- 1/s) i.e. P = [L/T(1/l- 1/s)] * T = 333 * 10^3 J/kg * (-1) K/[273 K (1/1000-1/917) m3/kg]

= 135 atm (approx)

(b) P= h * * g = 135 atm = 135 * 1.01325 * 10^5 N/m2 = h * 0.917 kg/m3 * 0.982 m/s2

i.e. h = [(135 * 1.01325 * 10^5)/(0.917 * 0.982)] m = 15.19 * 10^3 km

Hence, under 15.19 * 10^3 km distance of the glacier, the required pressure is observed.

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