Question

At 1 atm, how much energy is required to heat 75.0 g H2O(s) at −22.0 ∘C to H2O(g) at 147.0 ∘C? Use the heat transfer constants found in this table.

Answer 1

Following is the - complete Answer -&- Explanation: for the given: Question: in...typed format....

$\Rightarrow$Answer:

Required energy ( i.e. heat ): Q_total =  236.095 kJ ( kilo-joule, approx. ) , under STP...

$\Rightarrow$Explanation:

Following is the complete Explanation: for the above Answer.

• Given:

1. Pressure of the system: P = 1.0 atm.
2. Mass of Ice : m_ice = 75.0 g ( grams ) = mass of water as well, when melted: i.e. m_w
3. We know: specific heat capacity of Ice:  C_ice = 2.03 J / g . ^oC , at STP...
4. Initial temperature if Ice: T_i = - 22.0 ^oC
5. We know: latent heat of ice: $\Delta$ H_fus^o = 334.0 J / g , at STP...
6. We know: specific heat of water: C_w = 4.184 J / g. ^oC ( i.e. 1.0 g of H₂O, will require: 4.184 Joule of energy to increase its temperature by 1.0 ^oC )
7. We know: standard heat of vaporization of water: $\Delta$ H_vap^o = 2256.4 J / g ( i.e Joule per grams )
8. We know: standard specific heat of steam: C_steam = 2.01 J / g . ^oC

• ​​​​​​​Step - 1:

​​​​​​​We will get the following value of the heat required to raise the temperature of ice, from: - 22.0 ^oC to 0.0 ^oC, i.e. Q₁ ...

$\Rightarrow$Q₁ =    m_ice x  C_ice x  $\Delta$T =  ( 75.0 g ) x (2.03 J / g.^oC ) x ( 0.0 - ( - 22.0 ) ) ^oC = 3349.5 J ( Joule )

• Step - 2:

Using the given data: we will find: the value of the net amount of heat: required, to melt the ice, at 0.0 ^oC ; i.e. Q₂ ..

$\Rightarrow$ Q₂ =   m_ice x $\Delta$ H_fus^o =  ( 75.0 g ) x ( 334.0 J / g ) = 25050.0 J ( Joule )

• Step - 3:

​​​​​​​Using the given: data: we will get to find the following value of the heat required to raise the temperature of water, from:

0.0 ^oC to 100.0 ^oC, as Liquid water.... i.e. Q₃

$\Rightarrow$ Q₃ =  m_w x C_w x $\Delta$ T =  ( 75.0 g ) x ( 4.184 J / g. ^oC ) x ( 100 - 0 ) ^oC = 31380 J ( Joule )

• Step - 4:

​​​​​​​Using the given data: we will get the following value for the: heat required to change the phase of water, into steam, at 100 ^oC : i.e. ..Q₄ ...

$\Rightarrow$ Q₄ = m_w x  $\Delta$H_vap^o =  ( 75.0 g ) x ( 2256.4 J / g ) = 169230.0 J ( Joule )

• Step - 5:

​​​​​​​Using the given data: we can find the value of the heat required to raise the temperature of steam, from 100 ^oC, to 147.0 ^oC, ....as the following ( i.e. Q₅) :

$\Rightarrow$  Q₅ = m_steam x  C_steam x $\Delta$ T = ( 75.0 g ) x ( 2.01 J / g . ^oC ) x ( 147.0 - 100 ) ^oC = 7085.25 J ( Joule )

• Step - 6:

​​​​​​​Therefore: the total amount of heat required to fulfill the desired purpose ( i.e. Q_total  ) , will be the following:

$\Rightarrow$Q_total =  Q₁ + Q₂ + Q₃ + Q₄ + Q₅ =  ( 3349.5 J +  25050.0 J + 31380 J + 169230.0 J + 7085.25 J

= 236094.75 J

=  236.095 kJ ( kilo-joule, approx. )

• Answer:

​​​​​​​$\Rightarrow$Q_total =  236.095 kJ ( kilo-joule, approx. ) , under STP...