How much energy is required to convert 429 g of water to steam at 156 °C?

#C_text(water) = "4.184 J/(°C·mol)"; C_text(vapour) = "2.078 J/(°C·mol)"; Δ_text(vap)H = "2257 J/g"#

Answer 1

#~~1155(kJ)/(g*^oC)#.

We will be using the formula: #mcDeltaT#, where: => #m# is the mass of the object in grams. => #c# is the specific heat capacity of the object. => #DeltaT# is the change in temperature found by subtracting the initial temperature from the final.

Let's go over what we are given:

#m = 429 "g"#
#c_"liquid water" = 4.184 J/(g*^oC)#
#c_"water vapour" = 2.078 J/(g*^oC)#
# DeltaT = T_"final" - T_"initial"#
Now there's something we have to consider, we can't just get from liquid at #24^oC# to steam at #156^oC#. What we have to do is convert from liquid to it's boiling point of #100^oC#, and then from #100^oC# to #156^oC#.

So let's do the liquid to gas first.

#q_"liquid water" =mcDeltaT#
#=(429)(4.184)(100-24)#
#=136415.136 J/(g*^oC)#
Next, we convert the liquid at #100 °"C"# to vapour at #100 °"C"#. For this, we
#q_"conversion"=mDeltaH#
#= 429(2257)#
#=968253#

Now let's do the gas to hotter gas.

#q_"water vapour"=mcDeltaT#
#=(429)(2.078)(156-100)#
#=49921.87J/(g*^oC)#

Now all we have to do is add the energies.

#q_"total"=q_"liquid water" + q_"conversion" + q_"water vapour"#
#=136415.136+968253+49921.87#
#=1154590.01J/(g*^oC)#
#=1154.59001(kJ)/(g*^oC)#
#~~1155(kJ)/(g*^oC)#

Hope this helps :)

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Answer 2

To convert 429 g of water to steam at 156 °C, the energy required can be calculated using the formula:

[ Q = m \times C \times ΔT + m \times L ]

where:

  • ( Q ) is the energy required (in joules)
  • ( m ) is the mass of water (in grams)
  • ( C ) is the specific heat capacity of water (4.18 J/g°C)
  • ( ΔT ) is the change in temperature (in °C)
  • ( L ) is the latent heat of vaporization of water (2260 J/g)

Using the given values:

  • ( m = 429 ) g
  • ( C = 4.18 ) J/g°C
  • ( ΔT = (100 - 156) ) °C = -56 °C (since water is converting to steam, the temperature change is from 100 °C to 156 °C)
  • ( L = 2260 ) J/g

Substituting these values into the formula:

[ Q = 429 \times 4.18 \times (-56) + 429 \times 2260 ]

[ Q = -127097.52 + 970140 ]

[ Q = 843042.48 , \text{J} ]

Therefore, the energy required to convert 429 g of water to steam at 156 °C is approximately 843,042.48 joules.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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