Q. 3.41 - An Introduction to General, Organic, and Biological Chemistry

Question

Use the heat eyuatson to calculate the energy, in joules and calories, for each of the following (see Table 3.11):

a. to beat$25.0 \mathrm{g}$ of water from $12.5°\mathrm{C}$ to $25.7°\mathrm{C}$

b. to beat$38.0 \mathrm{g}$ of copper from $122°\mathrm{C}$ to $246{}^{\circ }\mathrm{C}$

c. lost when$15.0 \mathrm{g}$ of ethanol, ${\mathrm{C}}_2{\mathrm{H}}_8\mathrm{O}$, cools from $60.5{}^{\circ }\mathrm{C}$ to $-42.00°\mathrm{C}$

d. lost when$125 \mathrm{g}$ of iron cools from $118{}^{\circ }\mathrm{C}$to $55°\mathrm{C}$

Step-by-Step Solution

Answer

(a) As a result, the required heat is $330 Cal$
(b) As a result, the required heat is$434 Cal$
(c) As a result, the required heat is $904 Cal$
(d) As a result, the required heat is $850 Cal$

Step 1: Given data (part a)

(a) We are given the following information:

The mass of water is $25.0 \mathrm{g}$.

The specific heat $\left(SH\right.$) for water is$4.184 \mathrm{J}/\mathrm{g}{}^{\circ }\mathrm{C}$.

The initial temperature is $\left(T_{\text{initial }}\right)$ is $12.5° \mathrm{C}$.

The final temperature is$\left(T_{\text{final }}\right)$ is$25.7° \mathrm{C}$.

Using the following conversion factor, the heat lost can be converted from joules to calories:

Step 2: The temperature change(part a)

The temperature change$\left(\Delta T\right)$ can be calculated by using the following formula:

$\Delta T=T_{\text{firal }}-T_{\text{initial }}$

$={25.7}^{\circ } \mathrm{C}-{12.5}^{\circ } \mathrm{C}$

$=13.2°C$

The temperature change$\left(\Delta T\right)$ is $13.2{}^{\circ } \mathrm{C}$.

Step 3: The heat required (part a)

The heat equation is

Heat $=$mass$\times \Delta T\times SH$

We get by substituting the value in the above equation.

$\text{ Heat }=25.0 \mathrm{g}\times 13.2° \mathrm{C}\times 4.184 \mathrm{J}/\mathrm{g}°\mathrm{C}$

$=1380 J$

As a result, the required heat is$1,380 \mathrm{J}$

Step 4:T he heat obtained can be converted from joules into calories (part a)

Using the following conversion factor, the heat lost can be converted from joules to calories:

$1.00 \mathrm{cal}=4.184 \mathrm{J}$

$\frac{1.0 0\mathrm{cal}}{4.184 \mathrm{J}}$

The heat required is

Heat $=1,380 \mathrm{J}$

$=1,380 \mathrm{J}\times \frac{1.00 \mathrm{cal}}{4.184 \mathrm{J}}$

$=330 Cal$

As a result, the required heat is $330 cal$

Step 5:Given data (part b)

(b) We have the following information:

The mass of copper is $38.0 \mathrm{g}$.

The specific heat $( SH )$ for copper is $0.385 \mathrm{J}/\mathrm{g}°\mathrm{C}$

The initial temperature is $\left(T_{\text{initial }}\right)$ is $122° \mathrm{C}$.

The final temperature is $\left(T_{\text{final }}\right)$ is $246° \mathrm{C}$.

We must calculate the required heat in joules and calories..

Step 6:Temparature change (part b)

The temperature change $\left(\Delta T\right)$can be calculated by using the following formula:

$\mathrm{\Delta }T=T_{\text{final }}-T_{\text{initial }}$

$$\begin{gathered} =246° C-122° C \\ =124° C \end{gathered}$$

Step 7:The heat required (part b)

The heat equation is

$\text{ Heat }=\text{ mass }\times \Delta T\times SH$

We get by substituting the value in the above equation.

Heat $=38.0 \mathrm{g}\times 124° \mathrm{C}\times 0.385 \mathrm{J}/\mathrm{g}°\mathrm{C}$

$=1810 J$

As a result, the required heat is $1,810 \mathrm{J}$

Step 8:The heat obtained can be converted from joules into calories (part b)

Using the following conversion factor, the heat lost can be converted from joules to calories:

$1.00 \mathrm{cal}=4.184 \mathrm{J}$

$\frac{1 \mathrm{cal}}{4.184 \mathrm{J}}$

The heat required is

Heat $=1,810 \mathrm{J}$

$$\begin{gathered} =1,810 \mathrm{J}\times \frac{1.00 \mathrm{cal}}{4.184 \mathrm{J}} \\ =434 Cal \end{gathered}$$

As a result, the required heat is $434 \mathrm{cal}$

Step 9:Given data (part c)

(c) We have the following information:

The mass of ethanol is $15.0 \mathrm{g}$.

The specific heat$\left(SH\right.$) for ethanol is $2.46 \mathrm{J}/\mathrm{g}{}^{\circ }\mathrm{C}$

The initial temperature is$\left(T_{\text{initial }}\right)$ is $60.5° \mathrm{C}$.

The final temperature is$\left(T_{\text{final }}\right)$ is $42.0° \mathrm{C}$.

We have to calculate the heat released in joules and calories.

Step 10:The temperature change (part c)

The temperature change $\left(\Delta T\right)$ can be calculated by using the following formula:

$\mathrm{\Delta }T=T_{\text{final }}-T_{\text{initial }}$

$=-{42.0}^{\circ } \mathrm{C}-{60.5}^{\circ } \mathrm{C}$

$=-{102.5}^{\circ } \mathrm{C}$

Therefore, the temperature change $\left(\Delta T\right)$ is $102.5° \mathrm{C}$.

Step 11: The heat released (part c)

The heat equation is

$\text{ Heat }=\text{ mass }\times \Delta T\times SH$

We get by substituting the value in the above equation.

Heat $=15.0 \mathrm{g}\times \left(-102.5° \mathrm{C}\right)\times 2.46 \mathrm{J}/\mathrm{g}{}^{\circ }\mathrm{C}$

$=-3,780 \mathrm{J}$

The minus sign denotes that heat is lost during the process. As a result, the heat lost is$3,780 \mathrm{J}$.

Step 13:The heat released can be converted from joules into calories (part c)

Using the following conversion factor, the heat lost can be converted from joules to calories:

$1.00 \mathrm{cal}=4.184 \mathrm{J}$

$\frac{1 \mathrm{cal}}{4.184 \mathrm{J}}$

The heat released is

Heat $=3,780 \mathrm{J}$

$=3,780 J\times \frac{1.00 \mathrm{cal}}{4.184 \mathrm{J}}$

$=904 Cal$

As a result, the required heat is $904 \mathrm{cal}$

Step 13: Given data (part d)

(d) We are given the following information:

The mass of iron is $125 \mathrm{g}$.

The specific heat$S H$ for iron is $0.452 \mathrm{J}/\mathrm{g}°\mathrm{C}$

The initial temperature is$\left(T_{\text{initial }}\right)$ is $118° \mathrm{C}$.

The final temperature is ( $\left.T_{\text{final }}\right)$ is $55° \mathrm{C}$.

We have to calculate the heat lost in joules and calories.

Step 14: The temperature change (part d)

The temperature change$\left(\Delta T\right)$ can be calculated by using the following formula:

$\mathrm{\Delta }T=T_{\text{final }}-T_{\text{initial }}$

$={55}^{\circ }\mathrm{C}-{118}^{\circ } \mathrm{C}$

$=-{63}^{\circ }\mathrm{C}$

Therefore, the temperature change is $63° \mathrm{C}$.

The heat equation is

Heat$=$ mass $\times \Delta T\times SH$

We get by substituting the value in the above equation.

$\text{ Heat }=125 \underset{}{g}\times \left(-{63}^{\circ } \mathrm{C}\right)\times 0.452 \mathrm{J}/{\mathrm{g}}^{\circ }\mathrm{C}$

$=-3,600 \mathrm{J}$

The minus sign denotes that heat is lost during the process. As a result, the heat lost is $3,600 \mathrm{J}$

Step15: Converted from joules into calories (part d)

Using the following conversion factor, the heat lost can be converted from joules to calories:

$1.00 \mathrm{cal}=4.184 \mathrm{J}$

$\frac{1 \mathrm{cal}}{4.184 \mathrm{J}}$

The heat lost is

$\text{ Heat }=3,600 \mathrm{J}$

$=3,600 \mathrm{J}\times \frac{1.00 \mathrm{cal}}{4.184 \mathrm{J}}$

$=850 Cal$

As a result, the required heat is$850 \mathrm{cal}$