How Does Adding Heat Affect Pressure in a Sealed Steam Container?

AI Thread Summary
Adding heat to a sealed steam container increases the internal energy and pressure of the system. The initial conditions are 200°C with a pressure of 1554.9 KPa, and the final state is 350°C with a pressure of 16529 KPa. Since the container is rigid, no work is performed, leading to the conclusion that the heat added equals the change in internal energy, calculated to be 1759 KJ. To determine the final pressure and heat added, specific volumes and quality calculations are necessary. The discussion emphasizes the relationship between heat, internal energy, and pressure in a closed steam system.
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1. Homework Statement

Given: A sealed, rigid container (cannot change shape or size) containing steam at the shown conditions has heat added until it reaches the final shown state.

Determine: a) heat added, b) work performed, c) change in internal energy of the system, and d) final system pressure

2. Homework Equations

Q = W + Δu

Also given: T1= 200 deg C, mass = 10kg, volume = 1m^3 (for initial conditions)
T2 = 350 C (for final conditions)

The pressures I found in the thermo tables for the corresponding temperatures are:
P1 = 1554.9 KPa
P2 = 16529 KPa

3. The Attempt at a Solution

What I know is it's a saturated vapor system since it stated steam in the problem statement and closed system so no work is done.

From that I was able to answer b) W = 0 and c) Q = 0 + m(u1-u2) --> 10Kg(2594.2-2418.3) KJ/kg = 1759 KJ of internal energy. The u's I got from thermo tables.

The two I'm having a hard time with is a) the heat added to the system and d) final system pressure.
 
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You already found the answer to a. Look at what you put for c. You also already found the final pressure.
 
Also I can tell the quality is not going to be 100%. You can use the volume and mass to find the specific volume. Than use that specific volume and the specific volumes from the table for state 1 and 2 to find the quality.
 
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