Is the Change in Internal Energy Just 4513 J When No Work Is Done?

The system includes the pressure cooker, water, and steam. If 2.0 g of water is vaporized in a sealed pressure cooker with 4513 J of added heat, the change in the system's internal energy can be determined using the equation U = Q - W. Since the system is doing no work, the change in internal energy would be 4513 J. There is no other relevant information provided.
  • #1
yinnxz
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Homework Statement
The lid of a pressure cooker forms a nearly
airtight seal. Steam builds up pressure and
increases temperature within the pressure
cooker so that food cooks faster than it does
in an ordinary pot. The system is defined as
the pressure cooker and the water and steam
within it.

If 2.0 g of water is sealed in a pressure
cooker and then vaporized by heating, and
4513 J must be added as heat to completely
vaporize the water, what is the change in the
system’s internal energy?
Relevant Equations
U = Q - W
Since the system is doing no work, would it be just 4513 J? I don't think there is any other information to use
 
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  • #2
yinnxz said:
Homework Statement: The lid of a pressure cooker forms a nearly
airtight seal. Steam builds up pressure and
increases temperature within the pressure
cooker so that food cooks faster than it does
in an ordinary pot. The system is defined as
the pressure cooker and the water and steam
within it.

If 2.0 g of water is sealed in a pressure
cooker and then vaporized by heating, and
4513 J must be added as heat to completely
vaporize the water, what is the change in the
system’s internal energy?
Relevant Equations: U = Q - W

Since the system is doing no work, would it be just 4513 J? I don't think there is any other information to use
Correct
 
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