Chemistry question

Question

A 250. gram sample of ice has a temperature -20.0˚C. Give all answers in kilojoules.

a. Calculate the energy required to heat the ice to 0.0˚C.

b. Calculate the energy required to melt the ice.

c. Calculate the energy required to heat the water from 0.0˚C to 100.0˚C.

d. Calculate the energy required to boil the water.

e. Calculate the energy required to heat the steam to 120.0˚C.

f. Calculate the total energy required for all five steps.

Answer 1

To answer this, we need the following information about water:

Specific heat of ice = 2.09 J/g/deg

Specific heat of liquid water = 4.184 J/g/deg

Specific heat of steam = 1.89 J/g/deg

∆Hfusion = 334 J/g

∆Hvap = 2260 J/g

The general equations we will use are q = mC∆T and q = m∆H, where q is heat, m is mass, C is specific heat, ∆T is change in temperature and ∆H is change in enthalpy.

Step a: q = (250.0 g)(2.09 J/g/deg)(20 deg) = 10450 J = 10.45 kJ

Step b: q = (250.0 g)(334 J/g) = 83,500 J = 83.5 kJ

Step c: q = (250.0 g)(4.184 J/g/deg)(100 deg) = 104,600 J = 104.6 kJ

Step d: q = (250.0g)(2260 J/g) = 565,000 J = 565 kJ

Step e: q = (250.0 g)(1.89 J/g/deg)(20 deg) = 9450 J = 9.45 kJ

Sum of all steps = 10.45 + 83.5 + 104.6 + 565 + 9.45 = 773 kJ