PLEASE SHOW ALL WORK! 1) 65 g of ice at -17°C is mixed with 256 g of water at 29

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PLEASE SHOW ALL WORK!
1) 65 g of ice at -17°C is mixed with 256 g of water at 29.2°C. Calculate the final temperature of the mixture. Assume that no energy in the form of heat is transferred to the environment. (Heat of fusion = 333 J/g; specific heat capacities: ice = 2.06 J/g x K, liquid water = 4.184 J/g x K)
2) In the reaction of barium chloride with potassium phosphate, barium phosphate and potassium chloride are produced. If 50 mL of 0.80 M potassium phosphate and 50 mL of 0.30 M barium chloride reacts such that the temperature rises 2.1°C and the calorimeter absorbs 15.0 J/°C. What is the enthalpy of the reaction in kJ? What does the molar enthalpy of formation for barium phosphate in kJ? Density of solution is 1.02 g/mL and C_w=4.18 J/g°C PLEASE SHOW ALL WORK!
1) 65 g of ice at -17°C is mixed with 256 g of water at 29.2°C. Calculate the final temperature of the mixture. Assume that no energy in the form of heat is transferred to the environment. (Heat of fusion = 333 J/g; specific heat capacities: ice = 2.06 J/g x K, liquid water = 4.184 J/g x K)
2) In the reaction of barium chloride with potassium phosphate, barium phosphate and potassium chloride are produced. If 50 mL of 0.80 M potassium phosphate and 50 mL of 0.30 M barium chloride reacts such that the temperature rises 2.1°C and the calorimeter absorbs 15.0 J/°C. What is the enthalpy of the reaction in kJ? What does the molar enthalpy of formation for barium phosphate in kJ? Density of solution is 1.02 g/mL and C_w=4.18 J/g°C PLEASE SHOW ALL WORK!

2) In the reaction of barium chloride with potassium phosphate, barium phosphate and potassium chloride are produced. If 50 mL of 0.80 M potassium phosphate and 50 mL of 0.30 M barium chloride reacts such that the temperature rises 2.1°C and the calorimeter absorbs 15.0 J/°C. What is the enthalpy of the reaction in kJ? What does the molar enthalpy of formation for barium phosphate in kJ? Density of solution is 1.02 g/mL and C_w=4.18 J/g°C

Explanation / Answer

1)
heat required to take ice from -17 oC to 0 oC,
Q1 = Mi*Ci*delta Ti
= 65*1.06*(17)
= 1171.3 J

Heat required to take water from 29.2 oC to 0 oC,
Q2 = Mw*Cw*delta Tw
= 256*4.184*29.2
= 31276 J

heat required to melt whole ice,
Q3 = Mi*Li
= 65*333
= 21645 J

Q1+Q3 = 1171.3 + 21645 = 22816.3 J < Q2
So whole of ice will melt
let final temperature be ToC

Q1 + Q3 + heat required by melted ice = heat released by water
22816.3 + 65*4.184*T = 256*4.184*(29.2-T)
22816.3 + 267.8*T = 31276 - 1071.1*T
T = 6.3 oC
Answer: 6.3 oC

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