How much heat is required to change 48.5 g of liquid mercury (Hg) at 400 K to vapor at a 700 K? The boiling point of mercury is
2 answers:
Answer:
1.57 kJ
Explanation:
Thinking process:
The process of boiling mercury involves two heats:
1. heat required to bring the mercury to the boiling point.
2. heat required to maintain the boiling of the mercury
The specific heat capacity of mercury = 0.14 kJ/kgK
therefore, the heat will be:
= 0.0485×0.14×(629.88-400)
=1.56 kJ
The heat of vaporization = m
= 0.0485 × 0.29466
= 0.01429 kJ
the total energy = 1.56 + 0.01429
= 1.57 kJ
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Answer:
Energy= 2.7758 × 10^-11 J ;
71.112×10^-6 kJ.
Mass defect in Kilogram= 3.0885×10^-28 kg.
That is; 3.1×10^-28 kg(to two significant figure).
Explanation:
(Note: Check equation of reaction in the attached file/picture).
STEP ONE: we have to calculate the Mass defect.
Mass defect= Mass of reactants -- Mass of products.
Mass of the products: (140.9144+91.9262+3.060) u.
= 235.8666 u.
Mass of reactants: (1.0087+235.0439) u= 236.0526 u.
Therefore, the Mass defect= (236.0526 -- 235.8666) u
= 0.1860 u.
STEP TWO: Converting the Mass defect to energy;
0.18860 × 1.6605 × 10^27 kg
= 3.0885× 10^-28 kg
STEP THREE: Calculating energy released . Recall(from the question) c^2= 931.5 Mev/u. This is also equals to 9×10^16 m/s.
E=Mc^2.
Where E= energy released, c= speed of light, M= Mass.
Slotting in the values;
E= 3.0885×10^-28 kg × 9×10^16 m/s.
E=2.7758 × 10^-11 J.
Know that;( 1g of uranium × 1 mol of uranium ÷ 235.0439 g of uranium) × (6.002×10^23 atom of uranium/ 1 mol of uranium) × 2.7758× 10^-11.
=7.1112×10^-10 J
= 71.112×10^6 kJ.
