---
title: "Rigid container two-phase heated from 100 to 150°C"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 3
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:rigid-container-two-phase-100c-150c"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/rigid-container-two-phase-100c-150c
site: BCT Engineering
---

# Rigid container two-phase heated from 100 to 150°C

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

- Asked in: 2072 Magh, 8 marks
- Video: [Liquid water into Vapor, but volume stays constant  l Thermodynamics](https://www.youtube.com/watch?v=yRPnpazT5i8) — Engineering Hack
- 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-two-phase-100c-150c
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-3

## Question

A closed rigid container of volume 0.5 m³ is placed on a hot plate. Initially, the container holds two phase mixture of saturated liquid water and saturated water vapor at T1 = 100°C with a quality of 0.2. After heating, the temperature in the container is T2 = 150°C. Indicate the initial and final states on P-v and T-v diagrams and determine:
a) The pressure at each state
b) The mass of the vapor present at each state, in kg
c) If the heating continued, determine the temperature, when the container holds only saturated vapor. [Refer the attached tables for properties of steam]

## Short answer

At 100°C, pressure 101.4 kPa, vapor mass 0.298 kg. At 150°C, pressure 476.1 kPa, vapor mass 1.273 kg. Heating to saturated vapor at constant volume occurs at 156.4°C.

## Answer

**Given:** V = 0.5 m³ (rigid, constant v), state 1: T₁ = 100 °C, x₁ = 0.2; state 2: T₂ = 150 °C.

**Steam table values:**
| T | P_sat (kPa) | v_f | v_g (m³/kg) |
|---|---|---|---|
| 100 °C | 101.4 | 0.001043 | 1.6719 |
| 150 °C | 476.1 | 0.001091 | 0.3925 |

#### a) Pressures
P₁ = P_sat(100 °C) = **101.4 kPa (≈ 0.101 MPa)**  
State 2: v is constant. First find v₁:  
v₁ = v_f + x v_fg = 0.001043 + 0.2 × (1.6719 − 0.001043) = 0.3352 m³/kg  
At 150 °C, v_f < 0.3352 < v_g = 0.3925, so state 2 is still a wet mixture: P₂ = P_sat(150 °C) = **476.1 kPa**

#### b) Mass of vapour at each state
Total mass m = V/v₁ = 0.5/0.3352 = 1.4916 kg  
State 1: m_g1 = x₁ m = 0.2 × 1.4916 = **0.298 kg**  
State 2: x₂ = (v − v_f)/(v_g − v_f) = (0.3352 − 0.001091)/(0.3925 − 0.001091) = 0.8536  
m_g2 = 0.8536 × 1.4916 = **1.273 kg**

#### c) Temperature when only saturated vapour remains
Set v_g(T) = 0.3352 m³/kg. From the saturated tables: v_g = 0.3465 at 155 °C and 0.3068 at 160 °C. Interpolating:  
T = 155 + 5 × (0.3465 − 0.3352)/(0.3465 − 0.3068) ≈ **156.4 °C** (P_sat ≈ 565 kPa)

```text
 P|
  |  3   (sat. vapour, ≈ 565 kPa)
  |  2   (476.1 kPa)
  |  |
  |  |   all on v = 0.3352 = const
  |  1   (101.4 kPa)
  +----------------- v

 T|
  |  3   (≈ 156.4 C, on sat. vapour line)
  |  2   (150 C)
  |  |
  |  1   (100 C)
  +----------------- v
```
All three states lie on one vertical line (v = 0.3352). States 1 and 2 are inside the dome, and state 3 is on the saturated vapour line.

## Check yourself

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

1. How is the constant v found?
   A. v1 = vg at 100 C
   B. v1 = vf + x1 vfg at 100 C = 0.3352 m3/kg
   C. v1 = V x x1
   D. v1 = vf
   Answer: B. x1 = 0.2 so v1 = 0.001043 + 0.2 x 1.6709. (Key point: Initial specific volume)

2. Why is state 2 at 150 C a wet mixture?
   A. 0.3352 > vg
   B. T exceeds Tc
   C. vf < 0.3352 < vg = 0.3925 at 150 C
   D. P2 is above Pc
   Answer: C. Hence P2 = Psat(150 C) = 476.1 kPa. (Key point: State 2 phase)

3. How is vapour mass at each state found?
   A. mg = x V
   B. mg = V/vg
   C. mg = m (1 - x)
   D. m = V/v, then mg = x m: 0.298 kg and 1.273 kg
   Answer: D. m = 1.4916 kg and x2 = 0.8536. (Key point: Mass of vapour)

4. How is T found when only saturated vapour remains?
   A. Find T with vg(T) = 0.3352 m3/kg, about 156.4 C
   B. Find T with vf = 0.3352
   C. T = 150 C
   D. T = 100 C
   Answer: A. Interpolating the saturated table between 150 and 160 C. (Key point: Saturated vapour temperature)

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