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18 Dec 2019

How many grams of ice at -18.0C are needed to bring 10.5 kg of water at 90.0C and 37.0C? (Change in heat fusion = 6.01 KJ/mole) (Change in heat vapor= 54.2 kJ/mole) (Cp(s) = 37.1 J/molC) (Cp(l) = 75.3 J/ molC) (Cp(g)=248.1 J/molC)

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31 Dec 2019

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Please help me solve this :( Iâm not sure how they got the answer!

This problem exploits the path independence of state functions (i.e. enthalpy and internal energy in this case).

Suppose 18.0 grams of water ice at 200. K are suddenly placed in thermal contact with a large heating block maintained at a temperature of 330. K.

Part A

Assuming the external pressure is constant and P_ext=1.00 atm, calculate the enthalpy change ÎH of the water, when 18.0 grams of water ice at 200. K are irreversibly converted to liquid water at 330K. Also calculate the enthalpy change of the heating block as a result of this process and the enthalpy change of the universe as a result of this process. For ice Cp=37.7 J/K. For liquid water Cp=75.3 J/K. For water ÎH_fusion=6.01 kJ/mole. Assume these quantities are constant between T=200K and T=330K.

Part B

Using your results from Part A, calculate the work done in this process and calculate ÎU for the water, for the heating block, and for the universe. Assume the density of liquid water is 983.2 kg m^-3. Assume the density of ice is 916.7 kg m^-3.

Two 20.0 g ice cubes at -11.0 C are placed into 225g of water at 25.0 C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature of the water after all the ice melts.

Heat capacity of H2O(s) 37.7 J/(mol K)

Heat capacity of H2O(l) 75.3 J/(mol K)

Enthalpy of fusion of H2O 6.01 KJ/mol

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How many grams of ice at -18.0C are needed to bring 10.5 kg of water at 90.0C and 37.0C? (Change in heat fusion = 6.01 KJ/mole) (Change in heat vapor= 54.2 kJ/mole) (Cp(s) = 37.1 J/molC) (Cp(l) = 75.3 J/ molC) (Cp(g)=248.1 J/molC)

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