Question

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.

Answer #1

Q = heat change for conversion of ice at -10 ^{o}C to
ice at 0 ^{o}C + heat change for

conversion of ice at 0^{o}C to water at 0^{o}C +
heat change for conversion of water at 0^{o}C to water at
100 ^{o}C +heat change for conversion of water at 100
^{o}C to vapour at 100 ^{o}C+ heat change for
conversion of vapour at 100 ^{o}C to vapour at 113
^{o}C

Amount of heat required , Q = mcdt + mL + mc'dt + mL' + mc"dt"

= m(cdt + L + c'dt' + L' + c"dt" )

Where

m = mass of ice = 13.0 g

c'' = Specific heat of steam = 2.01 J/g degree C

c' = Specific heat of water = 4.186 J/g degree C

c = Specific heat of ice= 2.09 J/g degree C

L’ = Heat of Vaporization of water = 2260 J/g

L= Heat of fusion of ice = 334.9 J/g

dt’’ = 113-100 = 13^{o}C

dt' = 100 -0 =100 ^{o}C

dt = 0-(-10)=10 ^{o}C

Plug the values we get Q = m(cdt + L + c'dt '+ L' + c"dt" )

= 39.8 x10^{3} J

= 39.8 kJ

Therefore the amount of heat required is 39.8 kJ

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.

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.

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.

A) How much heat (in kJ) is required to warm 13.0 g of ice,
initially at -14.0 ∘C, to steam at 114.0 ∘C? The heat capacity of
ice is 2.09 J/g⋅∘C and that of steam is 2.01 J/g⋅∘C
B) How much heat is evolved in
converting 1.00 mol of steam at 130.0 ∘C to ice at -50.0 ∘C? The
heat capacity of steam is 2.01 J/(g⋅∘C) and of ice is 2.09
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 12.0 g of ice,
initially at -12.0 ∘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. kJ

How much heat in kilojoules is required to warm 10.0 g of ice,
initially at -10.0 ∘C, to steam at 125 ∘C. The heat capacity of ice
is 2.09 J/g∘C and that of steam is 1.84 J/g∘C.

5. How much heat (in kJ) is required to warm 10.0 g of ice,
initially at -10.0

How much heat is evolved in converting 1.00 mol of steam at
135.0 ∘C to ice at -50.0 ∘C? The heat capacity of steam is 2.01
J/(g⋅∘C) and of ice is 2.09 J/(g⋅∘C)

Please show each step. Thank you.
How much heat (in kJ) is evolved in converting 1.00 mole of
steam at 145.0 °C to ice at -50.0 °C? The heat capacity of ice is
2.09 J/g°C and that of steam is 2.09 J/g°C
Heat of fusion for water • Hfus = 6.02 kJ/mol
Heat of vaporization for water • Hvap = 40.7 kJ/mol

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