Question

1.

Use the following information on Cr to determine the amount of heat required to convert 186.3 g of solid Cr at 1760°C into liquid Cr at 2060°C.

melting point = 1860°C; boiling point = 2672°C

ΔH_fus = 20.5 kJ/mol; ΔH_vap = 339 kJ/mol;

c_(solid) = 44.8 J/g°C; c_(liquid) = 0.94 J/g°C

Enter your answer in units of kJ to three significant figures.

2.

Approximately how many ice cubes must melt to cool 700 milliliters of water from 29°C to 0°C? Assume that each ice cube contains 1 mole of H₂O and is initially at 0°C.

∆H(fusion) = 6.02 kJ/mol; ∆H(vaporization) = 40.7 kJ/mol

c(solid) = 2.09 J/g°C; c(liquid) = 4.18 J/g°C; c(gas) = 1.97 J/g°C

Enter your answer numerically.

Answer 1

Molar mass of Cr = 52 g/mol.

Moles of Cr = mass/molar mass

Heat required to increase temperature 1760 to 1860⁰c.

H1 = moles of Cr × specific heat of solid Cr × temperature change

Or, H1 = (186.3/52) × 44.8 × (1860 -1760)

Or H1 = (3.58 × 44.8 × 100) = 16038 J = 16.038 KJ.

Heat required to melt Cr

H2 = moles of Cr × heat of fusion of Cr

Or, H2 = 3.58 × 20.5 = 73.39 KJ.

Heat required to increase the temperature 2060⁰ c from 1860⁰c

H3 = 3.58 × specific heat of liquid Cr × temperature change

Or, H3 = 3.58× 0.94 × (2060 - 1860)

Or, H3 = 673 J = 0.673 KJ.

Now total heat needed

= H1 +H2 + H3

= 16.038 + 73.39 + 0.673

= 90.1 KJ. (3 significant figure)

2.

Let density of water = 1 g/mL

So, mass of water = 700 mL × 1 (g/mL)

Heat released by water = heat needed to melt ice.

Heat released to decrease the temperature of water from 29 to 0^oc.

q= m×C_+liquid) × T

Or, q = 700 × 4.18 × (29-0)

= 84854 J

= 84.854 KJ

Now

Heat needed to melt 1 mole ice

= Moles × heat of fusion per mol

= 6.02 KJ.

So, number of ice cube must melt to cool the water

= (84.854/6.02)

= 14.09

14.

.