---
title: "Rigid container of steam cooled to 90°C"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 3
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:rigid-container-200kpa-300c-cooled-90c"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/rigid-container-200kpa-300c-cooled-90c
site: BCT Engineering
---

# Rigid container of steam cooled to 90°C

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

- Asked in: 2072 Asoj, 8 marks
- Video: [Thermodynamics: First Law Rigid Tank. Steam Tables](https://www.youtube.com/watch?v=XQ1nlXe5ByI) — THE CLASS OF TORCHIA
- 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-200kpa-300c-cooled-90c
- 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 200 kPa, 300°C. It is cooled to 90°C. (Steam tables are attached here with)
a) At what temperature does a phase change start to occur?
b) What is the final pressure?
c) What mass fraction of the water is liquid in the final state?

## Short answer

Cooling at constant specific volume 1.3162 m³/kg. Condensation starts at 107.4°C. Final state at 90°C is a wet mixture with pressure 70.2 kPa and liquid mass fraction 0.442.

## Answer

**Given:** V = 0.170 m³ (rigid, so v is constant), state 1: 200 kPa, 300 °C, cooled to 90 °C.

**State 1 (superheated steam tables):** at 200 kPa, 300 °C: v₁ = 1.3162 m³/kg.  
Mass m = V/v₁ = 0.170/1.3162 = 0.1292 kg

The process is at constant volume, v = 1.3162 m³/kg.

#### a) Temperature at which the phase change starts
Condensation begins when v = v_g(T). From the saturated steam (temperature) table:
- 105 °C: v_g = 1.4185 m³/kg
- 110 °C: v_g = 1.2094 m³/kg

Interpolating for v_g = 1.3162:  
T = 105 + 5 × (1.4185 − 1.3162)/(1.4185 − 1.2094) = **≈ 107.4 °C** (P_sat ≈ 0.13 MPa)

#### b) Final pressure
At 90 °C, v = 1.3162 is less than v_g = 2.3591 and more than v_f = 0.001036, so it is a wet mixture; P = P_sat(90 °C) = **70.2 kPa** (≈ 0.07 MPa).

#### c) Mass fraction of liquid
x = (v − v_f)/(v_g − v_f) = (1.3162 − 0.001036)/(2.3591 − 0.001036) = 0.5577  
Liquid fraction = 1 − x = **0.442** (m_liquid ≈ 0.0571 kg)

```text
 P|
  |  1   (200 kPa, 300 C)
  |  |
  |  |   const v = 1.316
  |  |
  |  2   (70.2 kPa, wet)
  +----------- v

 T|
  |  1   (300 C)
  |  |
  |  *   (≈ 107.4 C, meets sat. vapour line)
  |  |
  |  2   (90 C, wet)
  +----------- v
```

**Answer: (a) ≈ 107 °C (b) 70.2 kPa (c) 44.2 % liquid by mass.**

## Check yourself

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

1. What stays constant in the cooling process?
   A. Pressure at 200 kPa
   B. Specific volume v = 1.3162 m3/kg
   C. Temperature
   D. Quality
   Answer: B. Rigid container, fixed mass, so v is constant. (Key point: Constant volume)

2. How is the temperature at which phase change starts found?
   A. Use Tsat at 200 kPa
   B. Use T = 90 C
   C. Find T where vg(T) = 1.3162 m3/kg, about 107 C
   D. Use vf = 1.3162
   Answer: C. Condensation begins when the constant v meets the saturated vapour line. (Key point: Start of condensation)

3. What is the final pressure at 90 C?
   A. 200 kPa
   B. 101.3 kPa
   C. 1.3162 kPa
   D. Psat(90 C) = 70.2 kPa since it is a wet mixture
   Answer: D. v lies between vf and vg at 90 C, so P = Psat. (Key point: Final pressure)

4. How is liquid mass fraction found?
   A. 1 - x with x = (v - vf)/(vg - vf), about 0.442
   B. x itself, about 0.558
   C. vf/vg
   D. (vg - v)/vg
   Answer: A. x = 0.558 so liquid fraction is 0.442. (Key point: Liquid fraction)

## Close variants set in other exams

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