Q. 3.50 - An Introduction to General, Organic, and Biological Chemistry

Question

Calculate the heat change at ${100}^{\circ }C$ for each of the following. and indicate whether heat was absorbed/released:

a. calories to condense $10.0 g$ of steam

b. joules to condense $7.60 g$ of steam

E. Kilocalories to vaporize $44 g$ of water

d. kilojoules to vaporize $5.00 Kg$ of water

Step-by-Step Solution

Answer

(a) The Calories to condense $10.0 g$ of steam is $540\times 10 cal$

(b) The Joules to condense $7.60 g$ of steam is $176\times {10}^2 J$

(c) The Kilocalories to Vaporize $44 g$ of water is $24 Kcal$

(d) The Kilojoules to Vaporize $5.00 Kg$ of water is $113\times {10}^2 KJ$

Step1: Introduction (part a).

A calorie is a type of energy unit. Calories are the energy people get from food and drink they consume, as well as the energy they use in physical activity.

Step2: Find the Calories (part a).

To condense $10.0 g$ of steam, we must calculate the quantity of heat absorbed or released at $100$degrees in calories.

The transition from a gas to a liquid form is known as condensation. Steam must release heat in order to produce this shift in condition.

Using the heat of condensation for steam, the amount of heat that has to be released can be estimated. The heat released to condense exactly $1 g$of steam at its condensation point is also known as condensation heat.

From Standards,

$$\begin{gathered} 1 g of H_{2}O=540 cal of heat \\ =\frac{540 cal}{1 g of H_{2}O} \end{gathered}$$

By the given information,

$$\begin{gathered} Heat=10.0 g of H_{2}O\times \frac{540 cal}{1 g of H_{2}O} \\ =5400 cal \\ =540\times 10 cal \end{gathered}$$

By the Condensation, the Total heat released is $540\times 10 Kcal.$

Step3: Introduction (part b).

In the International System of Units (SI), a joule is a unit of work or energy equal to the work done by a force of one newton acting through one meter.

To find Joules to condense $7.60 g$ of steam.

Heat change at ${100}^{\circ }C$

Step4: Find the Joules (part b).

To condense $7.60 g$of steam, we must calculate the quantity of heat absorbed or released at ${100}^{\circ }$in joules.

Condensation of a given amount of steam results in a state shift from gas to liquid. Steam must release heat in order to produce this shift in condition.

Using the heat of evaporation for steam, the amount of heat that has to be released can be estimated. The heat released to condensed exactly $1 g$of steam at its condensation point is also known as condensation heat.

As we know,

$$\begin{gathered} 1 g of H_{2}O=2260 J of heat \\ =\frac{2260 J}{1 g of H_{2}O} \end{gathered}$$

Using the Information given,

$$\begin{gathered} Heat=7.60 g of H_{2}O\times \frac{2260 J}{1 g of H_{2}O} \\ =17176 J \\ =176\times {10}^2 J \end{gathered}$$

By condensing the steam, the Heat released is $176\times {10}^2 J$

Step5: Introduction (part c).

A kilocalorie is another term for a calorie, so $1000$ calories will be written as $1,000kcals$.

Given information are:

To find Kilocalories to Vaporize $44 g$ of water

Heat change at ${100}^{\circ }C$

Step6: Find the Kilocalories (part c).

To evaporate $44 g$ of water, we must calculate the quantity of heat received or released at $100$degrees in kilocalories.

When a certain amount of water is vaporized, the condition of the water changes from liquid to gas. Heat should be absorbed by water to create this change in condition.

Using the heat of vaporization for water, the amount of heat that must be absorbed can be estimated. The heat absorbed to evaporate exactly $1 g$of water at its boiling point is also known as the heat of vaporization.

We Know that,

$$\begin{gathered} 1 g of H_{2}O=540 cal of heat \\ =\frac{540 cal}{1 g of H_{2}O} \end{gathered}$$

By the Data,

$$\begin{gathered} Heat=44 g of H_{2}O\times \frac{540 cal}{1 g of H_{2}O} \\ =23760 cal \end{gathered}$$

We know that,

$$\begin{gathered} 1 Kcal=1000cal \\ =\frac{1 Kcal}{1000 cal} \end{gathered}$$

Steam in Kilocalories is as follows:

$$\begin{gathered} Heat=23760 cal\times \frac{1 Kcal}{1000 cal} \\ =24 kcal \end{gathered}$$

By process of Vaporization, the Heat absorbed is $24 Kcal.$

Step7: Introduction (part d).

A kilojoule is a unit of energy measurement, similar to how kilometers measure distance.

To find the kilojoules to vaporize $5 Kg$ of water.

Heat Change at ${100}^{\circ }C$

Step8: Find the Kilojoules (part d).

To evaporate $5.00 kg$of water, we must calculate the quantity of heat received or released at $100$degrees in kilojoules.

When a certain amount of water is vaporized, the condition of the water changes from liquid to gas. Heat must be taken by water to create this change in condition.

Using the heat of vaporization for water, the amount of heat that must be absorbed can be estimated. The heat absorbed to evaporate exactly $1 g$of water at its boiling point is also known as the heat of vaporization.

By standards,

$$\begin{gathered} 1 g of H_{2}O=2260 J of heat \\ =\frac{2260 J}{1 g of H_{2}O} \end{gathered}$$

The Water is mentioned in kilogram. so,

$$\begin{gathered} 1 kg= 1000g \\ =\frac{1000 g}{1 Kg} \end{gathered}$$

Amount of Steam is,

$$\begin{gathered} =5.00 Kg\times \frac{1000 g}{1 Kg} \\ =5000 g \end{gathered}$$

By the Given Information,

$$\begin{gathered} Heat=5000 g of H_{2}O\times \frac{2260 J}{1 g of H_{2}O} \\ =11300000 J \end{gathered}$$

Then,

$$\begin{gathered} 1 KJ= 1000 J \\ =\frac{1 KJ}{1000 J} \end{gathered}$$

The Ice is calculated in Kilojoules is as:

$$\begin{gathered} Heat=11300000 J\times \frac{1 KJ}{1 J} \\ =113\times {10}^2 KJ \end{gathered}$$

By the process of condensation, the Heat absorbed is $113\times {10}^2 KJ$