Question

5. How many kJ of energy are required to change 1.00 m^3 of pure water by 1.0*C? Assume a perfect system. The specific heat of the water is 4.184 J/g*C. The density of water is 1.000 g/mL.

6. 2.500 grams of metal X (molar mass 65.39 g/mole) was reacted with 100.0 mL of a 1.500 M HCl solution in a coffee cup calorimeter. The temperature went from 12.50 *C to 40.50 *C. Determine the reaction enthalpy per mole of metal X. The specific heat of the solution is 4.184 J/ g*C. Assume a solution density of 1.00 g/mL and a perfect system.

7. 20.12 grams of butane, C4H10, was combusted with oxygen in a bomb calorimeter. The temperature of .500 kg of water went from 5.00 *C to 25.89 *C. The specific heat of the water is 4.184 J/g*C. Assume a solution density of 1.00 g/mL. Determine the heat (kJ) evolved per mole of butane. Assume a perfect bomb calorimeter.

Answer 1

Answer – 5) We are given, volume = 1.00 m³ , Δt = 1.0 ^oC

First we need to convert the volume m³ to cm³

We know,

1. m³ = 1.0*10⁶ cm³

1 cm³ = 1 mL

So, volume = 1.0*10⁶ mL

We are given density 1.00 g/mL

So mass of water = 1.0*10⁶ g

We know formula for calculating the heat

q = m*C* Δt

    = 1.0*10⁶ g * 4.184 J/g^oC*1.0^oC

    = 4.184*10⁶ J

     = 4.184*10³ kJ

6) We are given, volume = 100.0 mL , [HCl] =1.500 M , ti = 12.5^oC, tf = 40.5^oC

Mass of X = 2.500 g

Molar mass of X = 65.39 g/mole, density = 1.00 g/mL

So volume of solution = 100.0 mL = 100.0 g

Total mass of solution = 100.0+2.500 = 102.500 g

We know formula for calculating the heat

q = m*C* Δt

    = 102.5g * 4.184 J/g^oC*(40.5 – 12.5)^oC

    = 12008.08 J

= 12.0 kJ

We know, q = -ΔH

So, ΔH = -12.0

Now we need to calculate the moles of X

Moles of X = 2.500/65.39 g.mol^-1

                   = 0.0382 moles

So, enthalpy per mole of metal X = -12.0 kJ / 0.0382 mol

                                                       = 314 kJ/mol

7) We are given, ti = 5.0^oC, tf = 25.89^oC

Mass of butane = 20.12 g , mass of water = 0.500 kg = 500 g

density = 1.00 g/mL

Total mass of solution = 500+20.12 = 520.12 g

We know formula for calculating the heat

q = m*C* Δt

    = 520.12 g * 4.184 J/g^oC*(25.89 – 5.0)^oC

    = 45460.4 J

     = 45.46 kJ

Now we need to calculate the moles of butane

Moles of butane = 20.12/58.12 g.mol^-1

                          = 0.346 moles

So, heat per mole of metal butane = 45.46 kJ / 0.346 mol

                                                         = 131 kJ/mol