A) A volume of 125 mL of H2O is initially at room temperature (22.00 C). A chill

Question

A) A volume of 125 mL of H2O is initially at room temperature (22.00 C). A chilled steel rod at 2.00 C is placed in the water. If the final temperature of the system is 21.10 C , what is the mass of the steel bar? Answer to three significant values. Use the following values: specific heat of water = 4.18 J/(gC) specific heat of steel = 0.452 J/(gC)

B) A calorimeter contains 16.0 mL of water at 11.5 C . When 1.40 g of X (a substance with a molar mass of 73.0 g/mol ) is added, it dissolves via the reaction X(s)+H2O(l)X(aq) and the temperature of the solution increases to 25.5 C . Calculate the enthalpy change, H, for this reaction per mole of X. Assume that the specific heat of the resulting solution is equal to that of water [4.18 J/(gC)], that density of water is 1.00 g/mL, and that no heat is lost to the calorimeter itself, nor to the surroundings. Express the change in enthalpy in kilojoules per mole to three significant figures.

C) Consider the reaction C12H22O11(s)+12O2(g)12CO2(g)+11H2O(l) in which 10.0 g of sucrose, C12H22O11, was burned in a bomb calorimeter with a heat capacity of 7.50 kJ/C. The temperature increase inside the calorimeter was found to be 22.0 C. Calculate the change in internal energy, E, for this reaction per mole of sucrose. Express the change in internal energy in kilojoules per mole to three significant figures.

Explanation / Answer

a)

Given

Water

Volume = 125 ml

density = 1 g/ml

mass = Volume * density = 125 ml * 1 g/ml = 125 g

Intial temperature T1 = 22 C

final temperature T2 = 21.1 C

Cp = 4.184 J/g.C

Heat = m * Cp * ( T2 - T1) = 125 g * 4.184 J/g.C * ( 22 - 21.1) C = - 470.7 J

negative sign indicates heat is released

Metal rod

Intial temperature = 2 C

final temperature = 21.1 C

Heat = m * Cp * (T2 - T1) = same as heat lost by water = 470.7 J

m * 0.452 J/g.C * ( 21.1 - 2) C = 470.7 J

m = 54.52 g Answer

b)

Given

water = 16 ml

density = 1 g/ml

Mass of water (solvent) = 16 ml * 1 g/ml = 16 g

Intial temperature T1 = 11.5 C

mass of solute X = 1.4 g

mass of solution = 16 g + 1.4 g = 17.4 g

Cp = 4.184 J/g.C

final temperature T2 = 25.5 C

Heat from the solution = m * Cp * ( T2 - T1) = 17.4 g * 4.184 J/g.C * ( 25.5 - 11.5) C = 1019.22 J

this is energy released by 1.4 g of solute

Molar mass of solute = 73 g/mol

No. of moles = 1.4 g / 73 g/mol = 0.0192 moles

so Enthalpy = Heat / No. of moles = 1019.22 J / 0.0192 moles = 53145 J/mol = 53.145 kJ/mol Answer

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