Answer to Question #305098 in General Chemistry for Shane

Question

1. How much energy is required to change 2600 gram of water at 100˚C into steam at the same temperature?

2.Calculate the amount of energy (in kJ) needed to heat 346 gram of liquid water from 0 OC to 182 OC. Assume that the

specific heat of water is 4.184 J/g OC over the entire liquid range and the specific heat of steam is 1.99 J/g OC.

3.Calculate the heat released when 68.0 gram of steam at 124 OC is converted to water at 45 OC

Answer 1

Energy required : Q = 5,876,000 J

Further explanation

Change water to gas (phase change)

Q = mLv (vaporization / condensation)

Lv=latent heat of vaporization(for water = 2,260J/g)

Given

mass water = 2600 g

Required

Energy

Solution

Input the value :

Q = 2600 x 2260 J/g

Q = 5,876,000 J

In heating mass m = 346g of water from -10°C to 182°C there are the following processes involved:

(i) heating of ice from -10°C to 0°C

 ΔQ(1) = msΔT 

here we consider s is specific heat capacity of ice = 2.108J/gK

ΔT is change in temperature = 0°C-(-10°C) = 10°C

 ⇒ ΔQ(1) = 346×2.108×10= 7293.68 J

(ii) conversion of ice to water at 0°C

  ΔQ(2) = mL

here L is latent heat of fusion = 330 J/g

⇒ ΔQ(2) = 346×330 = 114180 J

(iii) heating of water from 0°C to 100°C

 ΔQ(3) = msΔT

 here we consider s as the specific heat capacity of water = 4.184 J/g

 ⇒ ΔQ(3) = 346×4.184×100 J = 144766 J

(iv) conversion of water to steam at 100°C

 ΔQ(4) = mL

 here L is latent heat of evaporation = 2260 J/g

⇒ ΔQ(4) = 346×2260 J = 781960 J

(v) heating of steam from 100°C to 182°C

  ΔQ(5)= msΔT

 here we consider s as specific heat capacity of steam = 1.99 J/g

 ⇒ ΔQ(5) = 346×1.99×82 J = 56460.28 J

Now the total heat required to heat water from -10°C to 182°C is:

   ΔQ(1) + ΔQ(2) + ΔQ(3) + ΔQ(4) + ΔQ(5) = ΔQ

  ⇒ ΔQ = 1104660 J

  ΔQ = 1.104 ×  J is the total heat required to heat water from -10°C to 182°C