Physical question-5?

1) Assuming no heat exchange with the surroundings, we have:
Heat gain by the ice = Heat loss by the water
Now, let's study:
Heat loss by the water when falling from 45°C to 0°C = 2 × 4.19 × 45 = 377.1 kJ
Heat gain by the ice when rising from -10°C to 0°C = 5 × 2.1 × 10 = 105 kJ
Heat gain by the ice when melting at 0°C = 5 × 334 = 1670 kJ
So we can see that the amount of heat loss by the water is NOT enough to melt all ice to water at 0°C but SUFFICIENT to raise the temperature of the ice to 0°C.
Therefore, the resulting is a mixture of ice and water at 0°C.
2A) For isothermal process:
pV = constant
So if the pressure is reduced to one-third of the original pressure:
(p/3)V' = pV
V' = 3V
hence the ratio of final volume to the initial volume is 3:1.
2B) For adiabatic process:
pVγ-1 = constant
So if the pressure is one-third of the original pressure:
(p/3)(V')γ-1 = pVγ-1
(V')2/3 = 3V2/3
(V'/V)2/3 = 3
V'/V = 33/2
hence the ratio of final volume to the initial volume is 33/2:1 or √27:1.

好勁啊!!!
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