Considering the melting of 3.7 kg of ice to liquid watcr. Thc hcat of melting at

Question

Considering the melting of 3.7 kg of ice to liquid watcr. Thc hcat of melting at 1 Cm and 0.9998 g/cm', respectivcly; and are rclatively indcpendent of prcssurc and tempcraturc. The spccific heat capacity of ice is 2.113 kJ K1 kg', and that of liquid water is 4.18 kJ K1 kg a. What are the AS, and G of melting at 0 °C, 1 atm? b. What is the volume change at 0 °C, 1 atm? c, what would be the G of melting at 50 atm and 0 °C? d. What would be thc AG at -20 °C and 1 atm? c. Do you cxpect the mclting tempcraturc to bc lower than or highcr than 0 °C at

Explanation / Answer

Ans a) 3.7 kg of ice means 3700/18 = 205.55 moles of ice.

deltaH = change in H = latent heat times no. of moles of ice= 205.55* 6.007= 1234.77 kJ

now delta Q/T = delta S => 1234770/273 = 4522.96 J/K

delta G = delta H - T(deltaS). => delta G = 0

And b) Volume of 3.7 kg water= 3700/0.9998 = 3700.74 cc Volume of 3.7 kg ice = 3700/0.917 = 4034.89 cc . So the volume change is 3700.74- 4034.89 = -334.15 cc that is the volume has decreased by this much.

ANS c) There will be no melting at 0 degree Celsius and 50 atm. The melting temperature will higher.

Ans d) At -20 degree Celsius, delta H = specific heat of ice * mass = 2.113*3.7 = 7.81kJ = delta G

Ans e) The melting temperature will higher at 50 atm. Because at higher pressure melting occurs at higher temperatures for example when we press together two ice cubes they stick to each other because while pressing the pressure among the adjacent surfaces increase this increases the melting point. The ice melts and in a moment again solidifies as the surrounding temperature which is of ice is much lower and hence they stick to each other.

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