Respuesta :
Answer:
The total heat required is 691,026.36 J
Explanation:
Latent heat is the amount of heat that a body receives or gives to produce a phase change. It is calculated as: Q = m. L
Where Q: amount of heat, m: mass and L: latent heat
On the other hand, sensible heat is the amount of heat that a body can receive or give up due to a change in temperature. Its calculation is through the expression:
Q = c * m * ΔT
where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the change in temperature (Tfinal - Tinitial).
In this case, the total heat required is calculated as:
- Q for liquid water. This is, raise 248 g of liquid water from O to 100 Celsius. So you calculate the sensible heat of water from temperature 0 °C to 100° C
Q= c*m*ΔT
[tex]Q=4.184\frac{J}{g*C} *248 g* (100 -0 )C[/tex]
Q=103,763.2 J
- Q for phase change from liquid to steam. For this, you calculate the latent heat with the heat of vaporization being 40 and being 248 g = 13.78 moles (the molar mass of water being 18 g / mol, then[tex]\frac{248 g}{18 \frac{g}{mol} } =13.78 moles[/tex] )
Q= m*L
[tex]Q=13.78moles*40.79 \frac{kJ}{mol}[/tex]
Q=562.0862 kJ= 562,086.2 J (being 1 kJ=1,000 J)
- Q for temperature change from 100.0 ∘ C to 154 ∘ C, this is, the sensible heat of steam from 100 °C to 154°C.
Q= c*m*ΔT
[tex]Q=1.99\frac{J}{g*C} *248 g* (154 - 100 )C[/tex]
Q=25,176.96 J
So, total heat= 103,763.2 J + 562,086.2 J + 25,176.96 J= 691,026.36 J
The total heat required is 691,026.36 J
The specific heat can be defined as the amount of heat required to raise the temperature of one gram of substance by one degree Celsius. The total heat of the reaction has been 691.029 kJ.
What is heat of vaporization?
The heat of vaporization is the amount of heat required to convert the liquid to the vapor state.
The water at 0 degree Celsius has been converted to the water at 100 degree Celsius. The 100 degree Celsius water vaporized to 100 degree steam. The 100 degree steam will be converted to the 154 degree Celsius.
The conversion of 0 degree Celsius water to 100 degree Celsius
[tex]Q_1=mc\Delta T[/tex]
Substituting the values of mass ([tex]m[/tex]), specific heat ([tex]c[/tex]), and change in temperature ([tex]\Delta T[/tex]):
[tex]Q_1=248\;\times\;4.184\;\times\;(100-0)\\Q_1=103,763.2\;\text J\\Q_1=103.7632\;\rm k\text J[/tex]
The amount of heat required to convert 100 degree Celsius water to 100 degree Celsius steam has been:
[tex]Q_2=mL[/tex]
Substituting the values of mass ([tex]m[/tex]), and heat of vaporization ([tex]L[/tex]):
[tex]Q_2=248\;\times\;40.79\\Q_2-562.0862\;\rm kJ[/tex]
The amount of heat required to convert 100 degree Celsius steam to 154 degree Celsius steam has been:
[tex]Q_3=mc\Delta T[/tex]
Substituting the values of mass ([tex]m[/tex]), heat of steam ([tex]c[/tex]), and change in temperature ([tex]\Delta T[/tex]):
[tex]Q_3=248\;\times\;1.99\;\times\;(154-100)\\Q_3=25,176.96 \;\text J\\Q_3=25.1796\;\rm kJ[/tex]
The total amount of heat in the reaction has been:
[tex]Q=Q_1+Q_2+Q_3[/tex]
Substituting the values for the total heat of the reaction:
[tex]Q=103.7632+562.0862+25.1796\;\rm kJ\\\textit Q=691.029\;kJ[/tex]
The total heat of the reaction for the conversion of water from 0 degree Celsius to 100 degree Celsius is 691.029 kJ.
Learn more about heat of vaporization, here:
https://brainly.com/question/15140455
