2. A sample of iron has the dimensions of 2.0 cm x 3.0 cm x
2 answers:
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Explanation:
The given data is as follows.
Volume of water = 0.25
Density of water = 1000
Therefore, mass of water = Density × Volume
=
= 250 kg
Initial Temperature of water () =
Final temperature of water =
Heat of vaporization of water () at
is 2133 kJ/kg
Specific heat capacity of water = 4.184 kJ/kg/K
As 25% of water got evaporated at its boiling point () in 60 min.
Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg
Heat required to evaporate = Amount of water evapotaed × Heat of vaporization
= 62.5 (kg) × 2133 (kJ/kg)
= kJ
All this heat was supplied in 60 min = 60(min) × 60(sec/min) = 3600 sec
Therefore, heat supplied per unit time = Heat required/time = = 37 kJ/s or kW
The power rating of electric heating element is 37 kW.
Hence, heat required to raise the temperature from to
of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)
= 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)
= 125520 kJ
Time required = Heat required / Power rating
=
= 3392 sec
Time required to raise the temperature from to
of 0.25
water is calculated as follows.
= 56 min
Thus, we can conclude that the time required to raise the temperature is 56 min.
Answer:
Reagent O₂ will be consumed first.
Explanation:
The balanced reaction between O₂ and C₄H₁₀ is:
2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O
Then, by reaction stoichiometry, the following amounts of reactants and products participate in the reaction:
- C₄H₁₀: 2 moles
- O₂: 13 moles
- CO₂: 8 moles
- H₂O: 10 moles
Being:
- C: 12 g/mole
- H: 1 g/mole
- O: 16 g/mole
The molar mass of the compounds that participate in the reaction is:
- C₄H₁₀: 4*12 g/mole + 10*1 g/mole= 58 g/mole
- O₂: 2*16 g/mole= 32 g/mole
- CO₂: 12 g/mole + 2*16 g/mole= 44 g/mole
- H₂O: 2*1 g/mole + 16 g/mole= 18 g/mole
Then, by reaction stoichiometry, the following mass quantities of reactants and products participate in the reaction:
- C₄H₁₀: 2 moles* 58 g/mole= 116 g
- O₂: 13 moles* 32 g/mole= 416 g
- CO₂: 8 moles* 44 g/mole= 352 g
- H₂O: 10 moles* 18 g/mole= 180 g
If 78.1 g of O₂ react, it is possible to apply the following rule of three: if by stoichiometry 416 g of O₂ react with 116 g of C₄H₁₀, 62.4 g of C₄H₁₀ with how much mass of O₂ do they react?
mass of O₂= 223.78 grams
But 21.78 grams of O₂ are not available, 78.1 grams are available. Since you have less mass than you need to react with 62.4 g of C₄H₁₀, <u><em>reagent O₂ will be consumed first.</em></u>