Determine how much energy is required to convert 35.0 g of water at 27.5 ºC to steam at 155.0 ºC?
[ specific heat of water = 4.184 J g-1 - ºC-1 , specific heat of steam = 1.92 J g-1 - ºC-1 , and
enthalpy of vaporization of water = 2260 J/g ]?
a.1.87 x 104 J
b.9.34 x 104 J
c.9.78 x 104 J
d.2.89 x 105 J
e.6.35 x 104 J
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Verified answer
There's three parts to the answer.
1. Energy required (Q) to raise the temperature from 27.5°C to 100°C (water).
Using the equation
Q = m x Cp x theta
where
Q = energy in J
m = mass in grams = 35.0g.
Cp = specific heat of water in J g-1 - ºC-1 = 4.184J/g/°C.
theta = change in temp in °C. = 100°C - 27.5 = 7..5°C.
So Q = 35.0 * 4.184 * 72.5 = 10616.9
2. Energy required (Q) to vaporise the water at 100°C to steam at 100°C.
Using the equation Q = m x Hvap
where
Q = energy in J
m = mass in g
Hvap = enthalpy of vaporization of water = 2260 J/g
So Q = m x Hvap = 35.0g x 2260J/g = 79100 J
3. Energy required (Q) to raise the temperature from 100°C to 155°C (steam).
Using the equation
Q = m x Cp x theta
where
Q = energy in J
m = mass in grams = 35.0g.
Cp = specific heat of steam in J g-1 - ºC-1 = 1.92J/g/°C.
theta = change in temp in °C. = 155°C - 100 = 55°C.
So Q = 35.0 * 1.92 * 55 = 3696 J
Total Energy required = 10616.9 + 79100 + 3696 = 93412.9 J = 9.34 X10^4 J