140 cal of energy are added to a container of air also has 30 Cal of work done to it. By how much did the internal energy of the air inside the container changes.
To solve this problem, we need to understand the relationship between energy and work done on a gas. According to the first law of thermodynamics, the change in internal energy of a system is equal to the heat added to the system (energy) minus the work done by the system:
ΔU = Q - W
Where ΔU is the change in internal energy, Q is the heat added, and W is the work done.
In this case, we are given that 140 cal of energy is added (Q = 140 cal) and 30 Cal of work is done (W = 30 Cal). The units of energy need to be converted to a consistent unit. 1 cal = 1/1000 Cal, so 140 cal = 0.14 Cal.
Substituting these values into the equation, we have:
ΔU = 0.14 Cal - 30 Cal
Simplifying, we get:
ΔU = -29.86 Cal
Therefore, the internal energy of the air inside the container changes by approximately -29.86 Cal.
To determine how much the internal energy of the air inside the container changes, we need to consider the first law of thermodynamics, which states that the change in internal energy (\(ΔU\)) of a system is equal to the heat added (\(Q\)) to the system minus the work done (\(W\)) by the system.
Mathematically, this can be represented as:
\(ΔU = Q - W\)
Given that 140 cal of energy are added to the air (\(Q = 140\, \text{cal}\)) and 30 Cal of work is done by the air (\(W = 30\, \text{Cal}\)):
First, we need to convert the units into the same dimension. Since 1 Cal = 1000 cal, we have:
\(W = 30 \times 1000\, \text{cal} = 30000\, \text{cal}\)
Now, we can substitute the values into the formula:
\(ΔU = 140\, \text{cal} - 30000\, \text{cal}\)
Simplifying the equation:
\(ΔU = -29860\, \text{cal}\)
Therefore, the internal energy of the air inside the container changes by -29860 cal.