Question

# Consider two 35-gram ice cubes, each initially at -5°C. One is dropped into a at container...

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).

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

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