One liter (also one kg) of water (liquid H2O) at T=300K is in an insulated chamb

Question

One liter (also one kg) of water (liquid H2O) at T=300K is in an insulated chamber. Three ice cubes, 50g each, at an initial temperature of 100K (partially chilled in liquid nitrogen) are inserted into the chamber. [Recall that for water, Tp 273K, Ta- 373K Other constants below.] a) Use a three phase diagram to explain what initially happens when the ice is added. 5. initially, the 1 kg of liquid water cools down from 200K while the 0.15 kg of ice warms up from 100K. Whether the water or ice reaches 273 first is initially unclear. b) Use energy-interaction diagrams to determine which will reach its phase change temperatures first, the water (freezing point) or the ice (melting point). Also, find the temperature of the other substance at that point To cool the water to 273 delta E_th (1 kg)(4.18 kJ/kgk)(-27K)--112.86 k To warm the ice to 273 delta E_th (0.15 kg)(2.05 kJ/kgk)(173K)- 53.20 k The ice melts first. Temperature of water when ice starts to rmelte 300K. (53.2 k)/I(4.18 k/gcX1 i T water 287K e) What is the final state (temperature and phase) of the combined system? Additional Enerry required to cool water to 273 11286 k-53.20 kT-59.66 kT Energy required to completely melt the ice (0.15 kg)(333.5 kJ kg) 50.0 kr Ice melts first. Final phase is all liquid water. Temperature of 1 kg of water when ice completely melted- 287-(50.0 4.18)-275K Final temperature = 2749 K

Explanation / Answer

Referring to your doubt ,the energy 333.5kJ is the energy required to change 1 kg of ice to 1 kg of water at freezing point i.e. 0C .This heat is called latent heat of fusion.This heat will only change the state but temperature will be same.So need to consider this energy also while calculating the total heat required

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