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"#
Let's go over what we are given:
So let's do the liquid to gas first.
Now let's do the gas to hotter gas.
Now all we have to do is add the energies.
Hope this helps :)
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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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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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