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How much energy (heat) is required to convert 248 g of water from 0oC to 154oC? Assume that the water begins as a liquid, that the specific heat of water is 4.184 J/g.oC over the entire liquid range, that the specific heat of steam is 1.99 J/g.oC, and the heat of vaporization of water is 40.79 kJ/mol

User Sen Sokha
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1 Answer

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Answer: The amount of heat required is 775.7 kJ

Step-by-step explanation:

The processes involved in the given problem are:


1.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\2.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(100^oC,373K)\\3.)H_2O(l)(100^oC,373K)\rightarrow H_2O(g)(100^oC,373K)\\4.)H_2O(g)(100^oC,373K)\rightarrow H_2O(g)(154^oC,427K)

Now, we calculate the amount of heat released or absorbed in all the processes.

  • For process 1:


q_1=m* L_f

where,


q_1 = amount of heat absorbed = ?

m = mass of water or ice = 248 g


L_f = latent heat of fusion = 334 J/g

Putting all the values in above equation, we get:


q_1=248g* 334J/g=84832J

  • For process 2:


q_2=m* C_(p,l)* (T_(2)-T_(1))

where,


q_2 = amount of heat absorbed = ?


C_(p,l) = specific heat of water = 4.184 J/g°C

m = mass of water = 248 g


T_2 = final temperature =
100^oC


T_1 = initial temperature =
0^oC

Putting all the values in above equation, we get:


q_2=248g* 4.184J/g^oC* (100-0)^oC=103763.2J

  • For process 3:


q_3=m* L_v

where,


q_3 = amount of heat absorbed = ?

m = mass of water or ice = 248 g


L_v = latent heat of vaporization =
40.79kJ/mol* (1000)/(18)=2266.1J/g (Conversion factor used: 1 kJ = 1000 J and molar mass of water = 18 g/mol)

Putting all the values in above equation, we get:


q_3=248g* 2260J/g=560480J

  • For process 4:


q_4=m* C_(p,g)* (T_(2)-T_(1))

where,


q_4 = amount of heat absorbed = ?


C_(p,g) = specific heat of steam = 1.99 J/g°C

m = mass of water = 248 g


T_2 = final temperature =
154^oC


T_1 = initial temperature =
100^oC

Putting all the values in above equation, we get:


q_4=248g* 1.99J/g^oC* (154-100)^oC=26650.1J

Calculating the total heat absorbed, we get:


Q=q_1+q_2+q_3+q_4


Q=[84832+103763.2+560480+26650.1]J=775,725.3J=775.7kJ

Hence, the amount of heat required is 775.7 kJ

User Jrarama
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