---
title: "Rigid container of steam 1500 kPa cooled to 100°C"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 3
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:rigid-container-1500kpa-200c-cooled-100c"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/rigid-container-1500kpa-200c-cooled-100c
site: BCT Engineering
---

# Rigid container of steam 1500 kPa cooled to 100°C

Fundamental of Thermodynamics & Heat Transfer · Chapter 3: Properties of Common Substances

- Asked in: 2071 Magh, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/rigid-container-1500kpa-200c-cooled-100c
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-3

## Question

A rigid container with a volume of 0.170 m³ is initially filled with steam at 1500 kPa, 200°C. It is cooled to 100°C. Determine
a) Total mass of the system
b) Final pressure
c) Mass fraction of the liquid in the final state
Also sketch the process on P-v and T-v diagrams.
[Refer the attached table for properties of steam]

## Short answer

For rigid container volume 0.170 m³, initial superheated steam at 1500 kPa, 200°C, v1=0.1324 m³/kg, mass m = V/v1 = 1.284 kg. Cooled to 100°C at constant volume, state becomes wet at saturation pressure 101.4 kPa; with v2=v1, quality x=0.0786, so liquid mass fraction = 1-x = 0.921.

## Answer

**Given:** V = 0.170 m³ (rigid); state 1: 1500 kPa, 200 °C; cooled to 100 °C.

**State 1:** T_sat(1500 kPa) = 198.3 °C, so steam at 200 °C is slightly superheated. Superheated table: v₁ = 0.1324 m³/kg.

#### a) Total mass
m = V/v₁ = 0.170/0.1324 = **1.284 kg**

#### b) Final pressure
v₂ = v₁ = 0.1324 (constant volume). At 100 °C: v_f = 0.001043, v_g = 1.6719 m³/kg; since v_f < v₂ < v_g, it is a wet mixture.  
P₂ = P_sat(100 °C) = **101.4 kPa (≈ 0.1 MPa)**

#### c) Mass fraction of liquid
x₂ = (0.1324 − 0.001043)/(1.6719 − 0.001043) = 0.0786  
Liquid fraction = 1 − x₂ = **0.921** (m_f = 1.183 kg; m_g = 0.101 kg)

```text
 P|
  |  1   (1500 kPa, 200 C)
  |  |
  |  |   const v = 0.1324
  |  |
  |  2   (101.4 kPa, wet)
  +----------- v

 T|
  |  1   (200 C)
  |  *   (≈ 198 C, meets sat. vapour line)
  |  |
  |  2   (100 C, wet)
  +----------- v
```
On both diagrams, process 1-2 is a vertical line going down from state 1 (superheated vapour) into the dome, crossing the saturated vapour line at about 198 °C and ending at 2 inside the dome.

**Answer: m = 1.284 kg; P₂ = 101.4 kPa; liquid mass fraction = 0.921.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is state 1 superheated?
   A. 200 C is below Tsat
   B. 200 C exceeds Tsat = 198.3 C at 1500 kPa
   C. x = 0.5
   D. P is critical
   Answer: B. v1 = 0.1324. (Key point: State 1)

2. m =
   A. V x v1
   B. v1/V
   C. V/v1 = 1.284 kg
   D. V/vg
   Answer: C. 0.170/0.1324. (Key point: Mass)

3. P2 =
   A. 1500 kPa
   B. 0
   C. Pc
   D. Psat(100 C) = 101.4 kPa since wet
   Answer: D. v is between vf and vg at 100 C. (Key point: Final pressure)

4. Liquid fraction =
   A. 1 - x2 = 0.921
   B. x2 = 0.0786
   C. 0.0786 / 0.921
   D. vf/v
   Answer: A. x2 = (0.1324 - 0.001043)/(1.6719 - 0.001043). (Key point: Liquid fraction)

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