Joules Conversion: From Ice at 0°C to Steam at 100°C Explained

In this article, we will explore the required joules to convert 223 grams of ice at 0°C to steam at 100°C. The process involves several steps, each requiring a distinct amount of energy. We will break down these calculations and understand the underlying principles.

Understanding the Problem

We need to convert 223 grams of ice at 0°C to steam at 100°C. The process involves three main steps:

Melting the ice to water at 0°C using the heat of fusion. Heating the water from 0°C to 100°C using the specific heat of water. Converting the water to steam at 100°C using the heat of vaporization.

Step-by-Step Calculations

Step 1: Melting the Ice to Water at 0°C

The first step involves melting ice. The heat of fusion for ice is 334,000 joules per kilogram. Here, we are dealing with 223 grams, or 0.223 kilograms.

Q_1 m times L_f

Substituting the values:

Q_1 0.223 text{ kg} times 334,000 text{ J/kg} 74,582 text{ J}

Step 2: Heating the Water from 0°C to 100°C

The next step involves heating the water from 0°C to 100°C. The specific heat capacity of water is 4,186 joules per kilogram per degree Celsius.

Q_2 m times c times Delta T

Here, (Delta T) is the temperature change, which is 100°C - 0°C 100°C.

Q_2 0.223 text{ kg} times 4,186 text{ J/kg·°C} times 100°C 93,363 text{ J}

Step 3: Converting Water to Steam at 100°C

The final step is converting water to steam. The heat of vaporization of water is 2,260,000 joules per kilogram.

Q_3 m times L_v

Substituting the values:

Q_3 0.223 text{ kg} times 2,260,000 text{ J/kg} 503,180 text{ J}

Total Energy Required

The total energy required is the sum of Q_1, Q_2, and Q_3:

Q_{text{total}} Q_1 Q_2 Q_3

Q_{text{total}} 74,582 text{ J} 93,363 text{ J} 503,180 text{ J} 671,125 text{ J}

Hence, the total energy required to change 223 grams of ice at 0°C to steam at 100°C is 671,125 joules.

Additional Thermal Processes Example

Let's consider another example where 200 grams of ice at 0°C is converted to steam at 100°C. We will calculate the total heat required.

Step 1: Melting the Ice to Water at 0°C (Q1)

Using the heat of fusion, (Q_1 m times L_f):

[Q_1 0.200 text{ kg} times 336,000 text{ J/kg} 67,200 text{ J}]

Step 2: Heating the Water from 0°C to 100°C (Q2)

Using the specific heat of water, (Q_2 m times c times Delta T):

[Q_2 0.200 text{ kg} times 4,200 text{ J/kg·°C} times 100 text{ °C} 84,000 text{ J}]

Step 3: Converting Water to Steam at 100°C (Q3)

Using the heat of vaporization, (Q_3 m times L_v):

[Q_3 0.200 text{ kg} times 2,260,000 text{ J/kg} 452,000 text{ J}]

Total Heat Required (QT)

[QT Q_1 Q_2 Q_3]

[QT 67,200 text{ J} 84,000 text{ J} 452,000 text{ J} 603,200 text{ J}]

Energy Released in Reverse Process

In the reverse process, where 200 grams of steam at 100°C is converted back to ice at 0°C, two processes occur:

The condensation of steam to water at 100°C (Qphase change). The cooling of water from 100°C to 0°C (Qcooling).

Phase Change Energy (Qphase change)

[Q_{text{phase change}} mL_v 0.200 text{ kg} times 2,260,000 text{ J/kg} 452,000 text{ J}]

Cooling Energy (Qcooling)

[Q_{text{cooling}} mcDelta T 0.200 text{ kg} times 4,200 text{ J/kg·°C} times 100 text{ °C} 84,000 text{ J}]

Total Energy Released (Ereleased)

[E_{text{released}} Q_{text{phase change}} Q_{text{cooling}} 452,000 text{ J} 84,000 text{ J} 536,000 text{ J} 53.6 text{ kJ}]

Thus, the required energy to convert 200 grams of ice at 0°C to steam at 100°C is 603,200 J, and the total energy released in the reverse process is 53.6 kJ.