Liquid sodium is being considered as an engine coolant. How many grams of liquid sodium (minimum) are needed to absorb 1.30 MJ o
f energy in the form of heat if the temperature of the sodium is not to increase by more than 10.0 °C? Use C P = 30.8 J / ( K ⋅ mol ) for Na ( l ) at 500 K
2 answers:
Answer:
97 000 g Na
Explanation:
The absortion (or liberation) of energy in form of heat is expressed by:
q=m*Cp*ΔT
The information we have:
q=1.30MJ= 1.30*10^6 J
ΔT = 10.0°C = 10.0 K (ΔT is the same in °C than in K)
Cp=30.8 J/(K mol Na)
If you notice, the Cp in the question is in relation with mol of Na. Before using the q equation, we can find the Cp in relation to the grams of Na.
To do so, we use the molar mass of Na= 22.99g/mol
Now, we are able to solve for m:
=97 000 g Na
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Answer:
The change in entropy is -1083.112 joules per kilogram-Kelvin.
Explanation:
If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy (), in joules per gram-Kelvin, by the following model:
(1)
Where:
- Mass, in kilograms.
- Specific heat of water, in joules per kilogram-Kelvin.
,
- Initial and final temperatures of water, in Kelvin.
If we know that ,
,
and
, then the change in entropy for the entire process is:
The change in entropy is -1083.112 joules per kilogram-Kelvin.