Q. 3.49 - An Introduction to General, Organic, and Biological Chemistry

Question

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

a. calories to vaporize $10.0$ g of water

b. joules to vaporize $5.00$ g of water

c. kilocalories to condense $8.0$ kg of steam

d. kilojoules to condense $175$ g of steam

Step-by-Step Solution

Answer

(a) The calories to Vaporize $10.0 g$ of water by Vaporization is $540\times 10 cal$

(b) The Joules to Vaporize $5.00 g$ of water during the vaporization of a given amount of water will be $113\times {10}^2 J$

(c) The Kilocalories to condense $8.0 Kg$ of steam will be $43\times {10}^{2 }cal$

(d) The kilojoules to Condense $175 g$ of steam is $396 KJ$

Step1: Given Information (part a).

The given information is as,

Mass of water$=10.0 g$

The heat vary at ${100}^{\circ }C$ to find out heat is released/absorbed.

Step2: Find the Calories to vaporize 10 . 0   g of water (part a).

To evaporate $10.0 g$ of water, we must calculate the quantity of heat received or emitted at $100$ degrees in calories.

Water undergoes a state transition from liquid to gas during vaporization. Heat must be absorbed by the 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.

According to the Standards,

$1 g of H_{2}O=540 cal of heat$

$=\frac{540\mathrm{cal}}{1 {\mathrm{gH}}_2\mathrm{O}}$

Therefore,

$Heat=10.0 g of H_{2}O\times \frac{540\mathrm{cal}}{1 {\mathrm{gH}}_2\mathrm{O}}$

$$\begin{gathered} =5400 cal \\ =540\times 10 cal \end{gathered}$$

As a result, the quantity of heat absorbed during the vaporization of a given amount of water will be $540\times 10 cal$

Step3: Given Information (part b).

The following are the given information,

Mass of water$=5.00 g$

The heat changes at ${100}^{\circ }C$ to find heat was absorbed/released.

Step4: Find the Joules to Vaporize 5 . 00   g of water (part b).

To evaporate $5.00 g$ grammes of water, we must calculate the quantity of heat absorbed or released at $100$ degrees in joules.

Water undergoes a state transition from liquid to gas during vaporization. Heat must 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.

According to the norms,

$1g of H_{2}O=2260 joules of heat$

$=\frac{2260 joules of heat}{1g of H_{2}O}$

By using values,

$Heat=5.00g of H_{2}O\times \frac{2260 J}{1g of H_{2}O}$

$$\begin{gathered} =11300 J \\ =113\times {10}^2 J \end{gathered}$$

As a result, the quantity of heat absorbed during the vaporization of a given amount of water will be $113\times {10}^2 J$

Step5: Given Information (part c)

The information is given as follows:

Mass of steam $=8.0 Kg$

The heat change at ${100}^{\circ }C$ to find heat absorbed/released.

Step6: Step3: Find the Kilocalories to Condense 8 . 0   K g of Steam (part c)

To condense $8.00 Kg$ of steam, we must calculate the quantity of heat absorbed or released at $100$ degrees in kilocalories.

The state of this volume of steam changes from gas to liquid during condensation. Water must release heat in order to achieve this shift in condition.

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

By the Standards,

$1g of H_{2}O=540 cal of heat$

$=\frac{540 cal}{1g of H_{2}O}$

By,

$1 Kg= 1000 g$

$=\frac{1000 g}{1 Kg}$

$$\begin{gathered} Amount of steam=8.00 kg\times \frac{1000 g}{1 Kg} \\ =8000 g \end{gathered}$$

The Heat of Condensation will be,

$$\begin{gathered} Heat=8000 g of H_{2}O\times \frac{540 cal}{1 g of H_{2}O} \\ =43,20,000 cal \end{gathered}$$

By the process of Condensation, the amount of heat that is released is $43,20,000 cal$

$1 Kcal=1000 cal$

Therefore,

$=\frac{1 Kcal}{1000 cal}$

The Steam in Kilocalories will be,

$$\begin{gathered} Heat=43,20,000 cal\times \frac{1 Kcal}{1000 cal} \\ =4320 cal \\ =43\times {10}^{2 }cal \end{gathered}$$

Step7: Given Data (part d)

The following information are given:

Mass of steam$=175 g$

Heat change at ${100}^{\circ }C$ to find absorbed/released

Step8: The kilojoules to Condense 175   g of Steam (part d)

To evaporate $175 g$ of steam, we must calculate the quantity of heat absorbed or released at ${100}^{\circ }$in kilocalories.

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

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

$$\begin{gathered} 1 g of H_{2}O=2260 joules of heat \\ =\frac{2260 J of heat}{1 g of H_2 O} \end{gathered}$$

By using given information,

$$\begin{gathered} Heat=175 g of H_{2}O\times \frac{2260 J}{1 g of H_{2}O} \\ =395500 J \end{gathered}$$

We know, $1 KJ= 1000 J$

Therefore,

$=\frac{1 KJ}{1000J}$

Then,

$$\begin{gathered} Heat= 3995500 J\times \frac{1 KJ}{1000J} \\ =396 KJ \end{gathered}$$