Question

Calculate the amount of energy necessary to warm 10.0 g of ice from 0.0° C to 137.0° C. Express your answer in kJ.

Do not enter units of measurement, do not enter the answer in scientific notation.

The specific heat is 4.184 J/g∙°C for water and 1.99 J/g∙°C for steam. ΔHvap is 40.79 kJ/mol and ΔHfus is 6.01 kJ/mol.

Answer #1

Ti = 0.0 oC

Tf = 137.0 oC

here

Lf = 6.01KJ/mol =

6010J/mol

Lets convert mass to mol

Molar mass of H2O = 18.016 g/mol

number of mol

n= mass/molar mass

= 10.0/18.016

= 0.5551 mol

Heat required to convert solid to liquid at 0.0 oC

Q1 = n*Lf

= 0.5551 mol *6010 J/mol

= 3335.9236 J

Cl = 4.184 J/g.oC

Heat required to convert liquid from 0.0 oC to 100.0 oC

Q2 = m*Cl*(Tf-Ti)

= 10 g * 4.184 J/g.oC *(100-0) oC

= 4184 J

Lv = 40.79KJ/mol =

40790J/mol

Heat required to convert liquid to gas at 100.0 oC

Q3 = n*Lv

= 0.5551 mol *40790 J/mol

= 22640.9858 J

Cg = 1.99 J/g.oC

Heat required to convert vapour from 100.0 oC to 137.0 oC

Q4 = m*Cg*(Tf-Ti)

= 10 g * 1.99 J/g.oC *(137-100) oC

= 736.3 J

Total heat required = Q1 + Q2 + Q3 + Q4

= 3335.9236 J + 4184 J + 22640.9858 J + 736.3 J

= 30897 J

= 31.0 KJ

Answer: 31.0 KJ

Calculate the amount of energy in kilojoules needed to change
207 g of water ice at −10 ∘C to steam at 125 ∘C. The following
constants may be useful: Cm (ice)=36.57 J/(mol⋅∘C) Cm (water)=75.40
J/(mol⋅∘C) Cm (steam)=36.04 J/(mol⋅∘C) ΔHfus=+6.01 kJ/mol
ΔHvap=+40.67 kJ/mol

Calculate the amount of energy in kilojoules needed to change
405 g of water ice at −10 ∘C to steam at 125 ∘C. The following
constants may be useful:
Cm (ice)=36.57 J/(mol⋅∘C)
Cm (water)=75.40 J/(mol⋅∘C)
Cm (steam)=36.04 J/(mol⋅∘C)
ΔHfus=+6.01 kJ/mol
ΔHvap=+40.67 kJ/mol

Calculate the amount of energy in kilojoules needed to change
225 g of water ice at −10 ∘C to steam at 125 ∘C. The following
constants may be useful: Cm (ice)=36.57 J/(mol⋅∘C) Cm (water)=75.40
J/(mol⋅∘C) Cm (steam)=36.04 J/(mol⋅∘C) ΔHfus=+6.01 kJ/mol
ΔHvap=+40.67 kJ/mol

1.Calculate the amount of energy (in kJ) necessary to convert
377 g of liquid water from 0 degree C to water vapor at 167 degress
C. The molar heat of vaporization (Hvap) of
water is 40.79 kJ/mol. The specific heat for water is 4.184 J/g
degrees C, and for steam is 1.99 J/g degrees C. (Assume that the
specific heat values do not change over the range of temperatures
in the problem.)
2.Which substance has the highest vapor pressure at...

How much heat (in kJ) is required to warm 13.0 g of ice,
initially at -10.0 ∘C, to steam at 109.0 ∘C? The heat capacity of
ice is 2.09 J/g⋅∘C and that of steam is 2.01 J/g⋅∘C.

How much heat (in kJ) is required to warm 12.0 g of ice,
initially at -10.0 ∘C, to steam at 111.0 ∘C? The heat capacity of
ice is 2.09 J/g⋅∘C and that of steam is 2.01 J/g⋅∘C.

How much heat (in kJ) is required to warm 13.0 g of ice,
initially at -10.0 ∘C, to steam at 113.0 ∘C? The heat capacity of
ice is 2.09 J/g⋅∘C and that of steam is 2.01 J/g⋅∘C.

12.67
How much heat (in kJ) is required to warm 10.0 g of ice,
initially at -13.0 ∘C, to steam at 112.0 ∘C? The heat capacity of
ice is 2.09 J/g⋅∘C and that of steam is 2.01 J/g⋅∘C.

How much heat (in KJ) is required to warm 13.0 g of ice,
initially at -10.0 *C, to steam at 113.0 degress C? The heat
capacity of ice is 2.09 J/g degress C and that of steam is 2.01 J/g
* C. Show work.

A quantity of ice at 0.0 °C was added to 33.6 g of water at 41.0
°C to give water at 0.0 °C. How much ice was added? The heat of
fusion of water is 6.01 kJ/mol, and the specific heat is 4.18
J/(g•°C). ______ grams

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 4 minutes ago

asked 26 minutes ago

asked 53 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 2 hours ago

asked 2 hours ago

asked 2 hours ago

asked 2 hours ago

asked 2 hours ago