Water's heat of fusion is 80. cal/g , its specific heat is 1.0calg⋅∘C, and its heat of vaporization is 540 cal/g . A canister is
filled with 330 g of ice and 100. g of liquid water, both at 0 ∘C . The canister is placed in an oven until all the H2O has boiled off and the canister is empty. How much energy in calories was absorbed?
1 answer:
Refer to the diagram shown below.
State 1:
The heat of fusion is required for the ice to become water.
The amount of heat required is
Q₁ = (330 g)*(80 cal/g) = 26,400 cal
State 2:
Sensible heat is required to raise water at 0 C to the boiling point of 100 C.
The heat required is
Q₂ = (430g)*(1 cal/(g-C))*(100 C) = 43,000 cal
State 3:
The heat of vaporization is required to boil off the water.
The heat required is
(430 g)*(540 cal/g) = 232,200 cal
Total heat absorbed is
Q₁+Q₂+Q₃ = 301,600
Answer: 301,600 cal or 301.6 kcal
State 1:
The heat of fusion is required for the ice to become water.
The amount of heat required is
Q₁ = (330 g)*(80 cal/g) = 26,400 cal
State 2:
Sensible heat is required to raise water at 0 C to the boiling point of 100 C.
The heat required is
Q₂ = (430g)*(1 cal/(g-C))*(100 C) = 43,000 cal
State 3:
The heat of vaporization is required to boil off the water.
The heat required is
(430 g)*(540 cal/g) = 232,200 cal
Total heat absorbed is
Q₁+Q₂+Q₃ = 301,600
Answer: 301,600 cal or 301.6 kcal
You might be interested in
Answer:
Explanation:
Hello,
In this case, the enthalpy of combustion is understood as the energy released when one mole of fuel, in this case octene, is burned in the presence of oxygen and is computed with the enthalpies of formation of the fuel, carbon dioxide and water as shown below (oxygen is circumvented as it is a pure element):
Thus, since we already know the enthalpy of combustion of the fuel, for carbon and water we have -393.5 and -241.8 kJ/mol respectively, thereby, the enthalpy of combustion turns out:
Best regards.