Q. 3.54 - An Introduction to General, Organic, and Biological Chemistry

Question

Using the values for the heat of fusion, specific heat of water. and/or heat of vaporization, calculate the amount of heat energy in each of the following:

a. joules released when $125 g$ of steam at ${100}^{\circ }C$ condenses and cools to liquid at $15.0^{\circ }C$ 

b. kilocalories needed to melt a $525-g$ ice sculpture at $0^{\circ }C$ and to warm the liquid to $15.0^{\circ }C$ 

c. kilojoules released when $85.0 g$ of steam condenses at ${100}^{\circ }C$, cools, and freezes at $0^{\circ }C$

Step-by-Step Solution

Answer

(a) Joules released when $125 g$ of steam at ${100}^{\circ }C$ and cools at $15.0^{\circ }C$ is $3.28\times {10}^5 J$

(b) Kilocalories needed to melt $525 g$ of ice at $0^{\circ }C$and warms at $15.0^{\circ }C$ is $5.0\times {10}^4 cal$

(c) Kilojoules released when $85 g$ of steam condenses at ${100}^{\circ }C$ and freezes at $0^{\circ }C$ is $2.56\times {10}^2 KJ$

Step1: Given Information (part a).

The given information are used to find the joules:

Mass of steam$=125 g$

Initial temperature$={100}^{\circ }C$

Cooling temperature$=0^{\circ }C$

Step2: Find the joules when steam cooled (pat a).

Steam has a mass of $125 g$and a vaporization heat of $2260 J/g$ As a result, the formula for calculating the energy released to condense $125 g$of steam is:

$$\begin{gathered} Heat=mass\times heat of vaporization \\ =125 g\times \frac{2260 J}{g} \\ =2.83\times {10}^5 J \end{gathered}$$

Calculate the Heat energy,

$$\begin{gathered} Heat=m\times ∆t\times SH \\ =125 g\times \left({85}^{\circ }C-0^{\circ }C\right)\times \frac{4.184 J}{g^{\circ }C} \\ =4.45\times {10}^4 J \end{gathered}$$

The total energy is equal to the sum of the energy is released during condensation and the energy required to chill the liquid. The formula is as follows:

$$\begin{gathered} =2.83\times {10}^5 J+4.45\times {10}^4 J \\ =3.28\times {10}^5 J \end{gathered}$$

Then the Total amount of heat energy in Joule is $3.28\times {10}^5 J$

Step3: Given Information (part b).

The following are the information to find Kilocalories:

Mass of Ice$=525 g$

Initial temperature$=0^{\circ }C$

Heating temperature$={15}^{\circ }C$

Step4: Find the Kilocalories needed to melt and warm (part b).

The heat of fusion of water is $80 cal/g$, while the mass of ice is $525 g$. As a result, the calculation for calculating the amount of energy required to melt $525 g$of ice is as follows:

$$\begin{gathered} Heat=mass\times heat of fusion \\ =525 g\times \frac{80 cal}{g} \\ =4.20\times {10}^4 cal \end{gathered}$$

Calculate the Heat energy to warm the liquid from $0^{\circ }C$ to ${15}^{\circ }C$is

$$\begin{gathered} Heat=m\times ∆t\times SH \\ =525 g\times \left(15.0^{\circ }C-0^{\circ }C\right)\times \frac{1 cal}{g^{\circ }C} \\ =7.88\times {10}^3 cal \end{gathered}$$

The total energy is equal to the sum of the energy required to melt and warm the liquid. The formula is as follows:

$$\begin{gathered} =4.20\times {10}^4 cal+7.88\times {10}^3 cal \\ =5.0\times {10}^4 cal \end{gathered}$$

Therefore, the Total amount of heat in calories is $5.0\times {10}^4 cal$

Step5: Given Information (part c).

The following are the information to find Kilojoules is as:

Mass of steam$=85.0 g$

Initial temperature$={100}^{\circ }C$

Cooling temperature$=0^{\circ }C$

Step6: Find the Kilojoules released (part c).

Steam's mass is $85 g$. the formula to find the Heat released during condensation is as:

$$\begin{gathered} Heat=mass\times heat of vaporization \\ =85.0 g\times \frac{2260 J}{g} \\ =1.92\times {10}^5 J \end{gathered}$$

Calculate the Heat energy to cool down the liquid from ${100}^{\circ }C$to $0^{\circ }C$is as follows:

$$\begin{gathered} Heat=m\times ∆t\times SH \\ =85.0 g\times \left({100}^{\circ }C-0^{\circ }C\right)\times \frac{4.184 J}{g^{\circ }C} \\ =3.56\times {10}^4 J \end{gathered}$$

The following is the formula for calculating the amount of energy required to freeze the liquid:

$$\begin{gathered} Heat=mass\times heat of fusion \\ =85.0 g\times \frac{334 J}{g} \\ =2.84\times {10}^4 J \end{gathered}$$

The Total Energy calculated as follows:

$$\begin{gathered} 1.92\times {10}^5 J+3.56\times {10}^4 J+2.84\times {10}^4 J=2.56\times {10}^5 J\times \frac{1 KJ}{1000 J} \\ =2.56\times {10}^2 KJ \end{gathered}$$