My question is a General Chemistry that involves the heat capacity, heat of fusion, and heat of vaporization.

Question

Calculate each of the following:

 

a. the amount of heat required to warm 3.47 moles of water from 25*C to 55*C.

b. the mass of steam required to release 3.65 x 10 to the 5th power kJ on condensation.

c. the amount of heat required to melt 12.00kg of ice at 273.15K.

d. the volume of water that can be heated from 20.00*C to 30.0*C with 7.32kJ of heat. (Use the density at 25 degrees C).

e. the total amount of heat required to warm 5.00kg of ice from -25.0*C to 0.0*C and then melt it.

Answer 1

Ivy E. You didn't care for my previous answer to this identical question? How come? I submit the same answer below since I don't know what the problem is. Let me know so I can adjust and maybe we all can leaern something. Thanks.

a). q = mC∆T

q = heat = ?

m = mass = 3.47 mols H2O x 18 g / mol = 62.46 g H2O

C = specific heat = 4.184 J/gº

∆T = change in temperature = 55º - 25º = 30º

q = (62.46 g)(4.184 J/gº)(30º) = 7840 J = 7.84 kJ

b). q = m∆Hvap

q = heat = 3.65x10⁵ kJ

∆Hvap = heat of vaporization = 2260 kJ / kg

m = mass = ?

3.65x10⁵ kJ = (m)(2260 kJ/kg)

m = 161.5 kg = 162 kg

c). q = m∆Hf

q = heat = ?

m = mass = 12.00 kg

∆Hf = heat of fusion = 334 kJ/kg

q = (12.00 kg)(334 kJ/kg)

q = 4008 kJ

d). q = mC∆T

q = heat = 7.32 kJ = 7320 J

m = mass = ?

C = specific heat = 4.184 J/gº

∆T = change in temperature = 30º - 20º = 10º

7320 J = (m)(4.184 J/gº)(10º)

m = 175 g

e) to rasie temperature of the ice from -25º to 0º:

q = mC∆T

q = heat = ?

m = mass = 5.00 kg = 5000 g

C = specific heat of ice = 2.09 J/gº

∆T = change in temperature = 25º

q = (5000 g)(2.09 J/gº)(25º)

q = 261,250 J = 261.3 kJ

to melt the ice @ 0ºC:

q = m∆Hf

q = heat = ?

m = mass = 5000 g

∆Hf = heat of fusion = 334 J/g

q = (5000 g)(334 J/g)

q = 1670000 J = 1670 kJ

Total heat required = 261 kJ + 1670 kJ = 1930 kJ