Answer:
The half-life of polonium-210 is approximately 138.792 days.
Explanation:
We must remember that the decay of a radioisotope is modelled by this ordinary differential equation:
Where:
- Current mass of the isotope, measured in grams.
- Time constant, measured in days.
Whose solution is:
Where 
is the initial mass of the isotope, measured in grams.
Our first step is to determine the value of the time constant:
If we know that 
and 
, then the time constant of the radioisotope is:
And lastly we find the half-life of polonium-210 (
), measured in days, by using this expression:
The half-life of polonium-210 is approximately 138.792 days.
Answer:
Water is a good heat transfer fluid as it has a high thermal capacity and low viscosity. It is cheaper to use because its application in direct steam generation saves cost in the heat exchanger. However, it is unstable and difficult to manage at high temperatures.
Explanation:
hope this helps
Answer:
<h2>36 atm </h2>
Explanation:
The pressure required can be found by using the formula for Boyle's law which is
Since we are finding the required pressure
From the question we have
We have the final answer as
<h3>36 atm</h3>
Hope this helps you
Answer:
17 kJ
Explanation:
Calculation for the Calculate the energy required to heat 0.60kg of ethanol from 2.2°C to 13.7°C.
Using this formula
q = mC∆T
Where,
q represent Energy
m represent Mass of substance=0.60kg=600g
C represent Specific heat capacity=2.44J·g−1K−1.
∆T represent change in Temperature=2.2°C to 13.7°C.
Let plug in the formula
q=(0.60 kg x 1000 g/kg)(2.44 J/gº)(13.7°C-2.2°C)
q = (600g)(2.44 J/gº)(11.5º)
q=16.836 kJ
q= 17 kJ (Approximately)
Therefore the energy required to heat 0.60kg of ethanol from 2.2°C to 13.7°C will be 17 kJ
