Question

How much heat energy, in kilojoules, is required to convert 32.0 g of ice at −18.0 ∘C to water at 25.0 ∘C ? Express your answer to three significant figures and include the appropriate units.

How long would it take for 1.50 mol of water at 100.0 ∘C to be converted completely into steam if heat were added at a constant rate of 25.0 J/s ?

Express your answer to three significant figures and include the appropriate units.

Answer 1

1)

mass of Ice = 32.0g

a) Temperature changes from - 18C to 0C

m= 32.0 grams

T = 0 - ( -18) = 18C

specific heat capacity of ice = s= 2.09 J/gC

q1= msT

q1= 32.0 x 2.09 x 18= 1203.84 J

q2) phase change at 0C

m= 32 g

Hfusion = 336 J

q2= m x Hfus = 32.0 x 336 = 10752 J

q3)

temperaute changes from 0C to 25C

m= 32 g

T= 25-0 = 25C

specific heat capacity of water = 4.184 J/gC

q3= msT = 32 x 4.184 x 25= 3347.2 J

Total energy required = q1+q2+q3

Total energy = 1203.84 + 10752 + 3347.2 = 15303.07 J

Total energy = 15.303 KJ

Total energy = 15.3 KJ.

2)

number of moles of water = 1.50 moles

molar mass of water = 18.0gram/mole

number of moles = mass/molar mass

mass ofwater = number of moles x molar mass

mass of water = 1.50x 18.0 = 27.0 grams

heat of vaporisation of water = Hvap = 2250 J/g

Q= mx Hvap = 27.0 x 2250= 60750 J

rate = 20 J/s

time required = Q/rate = 60750J/20J/s = 3037.5 sec.

Time = 50.625 min

Time = 50.6 min.