Question

Consider two 35-gram ice cubes, each initially at -5°C. One is dropped into a at container filled with 0.1 kg of liquid nitrogen (initially the boiling point of nitrogen); the other cube is dropped into a container filled with 1 kg of (liquid) water, initially at 5°C. Assume that no heat is lost through either container (or into the atmosphere)

a) Describe what happens to each ice cube (and why)

b) If (Lv)Nitrogen =48 kcal/kg, (LF)Nitrogen = 6.1 kcal/kg, and cNitrogen =0.15 kcal/(kg °C), determine how much heat would be needed to produce 100 grams of nitrogen gas at 20°C.

c).Determine the final (equilibrium) temperature of the icewater mixture.

d) Determine the phase (ie., liquid or solid) corresponding to the equilibrium temperature you obtained in part (c).

Answer 1

a) Boiling point of nitrogen is -195.8 C, and the ice cube is at -5 C, so the ice cube's temperature drops to further lower level and there is no change of state pf the ice cube when put in boiling noitrogen
Water is at 5 C, ice cubes are at -5C so the ice cubes start to melt when put in water
b) heat = Lv*m + mcdT = 48*0.100 + 0.15*0.1*(20-(-195.8)) = 8.037 kCal

c) initial temperature of water, tw = 5 C
final temperature = T
initial mass of water, m = 1 kg
initaial temperature of ice = -5C = ti
initial mass of ice . M = 0.035 kg
hence from heat conservation'
energy lost by ice energy gained by water
MLf * Mc(ti) + MC(T - ti) = mC(tw - T)
here, Lf is latent heat of fusion of water, c is heat capacity of ice, C is heat capacity of water
0.035(333550 + 2108*5 + 4187(T + 5)) = 1*4187(5 - T)
T = 1.8827 C
d) Final phase = liquid