A person heats up a can of soup on a stove. If he does 250 joules of work on the soup by stirring it as the burner adds 1675 joules of heat, what is the change in its internal energy?
I took 1675j - 250 j = 1425j
To determine the change in the internal energy of the soup, you need to consider the energy transferred as heat (referred to as Q), any work done on or by the system (referred to as W), and the change in internal energy (referred to as ΔU).
In this case, the heat added to the soup is 1675 joules (Q = 1675 J), and the work done on the soup by stirring is -250 joules (W = -250 J).
The change in internal energy (ΔU) can be calculated using the first law of thermodynamics:
ΔU = Q - W
Substituting the given values:
ΔU = 1675 J - (-250 J)
ΔU = 1675 J + 250 J
ΔU = 1925 J
Therefore, the change in the internal energy of the soup is 1925 joules.
To determine the change in internal energy of the soup, we need to consider the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.
In this case, the heat added to the soup is 1675 joules and the work done on the soup is -250 joules (negative because work is being done on the soup). Therefore, the change in internal energy can be calculated as:
Change in internal energy = Heat added - Work done
Change in internal energy = 1675 J - (-250 J)
Change in internal energy = 1675 J + 250 J
Change in internal energy = 1925 J
So, the change in internal energy of the soup is 1925 joules.