Q. 3.42 - An Introduction to General, Organic, and Biological Chemistry

Question

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

a. to heat $5.25 \mathrm{g}$ of water from $5.5°\mathrm{C}$to $64.8{}^6\mathrm{C}$

b. lost when $75.0 \mathrm{g}$of water cools from$86.4°\mathrm{C}$to $2.1°\mathrm{C}$

c. to heat $10.0 \mathrm{g}$ of silver from$112°\mathrm{C}$ to $275°\mathrm{C}$

d. lost when$18.0 \mathrm{g}$ of gold cools from$224°\mathrm{C}$ to $118°\mathrm{C}$

Step-by-Step Solution

Answer

(part a) As a result, the required heat is 311 cal.

(part b) As a result, the heat lost is$633\times {10}^{\mathrm{\prime }} \mathrm{cal}$

(part c) As a result, the required heat is$91.6 \mathrm{cal}$

(part d) As a result, the heat lost is$58.8 \mathrm{cal}$

Step 1: Given data (part a)

(a) The mass of water is$5.25 \mathrm{g}$, and the specific heat $\left(SH\right)$ for water is $4.184 \mathrm{J}/\mathrm{g}°\mathrm{C}$ The initial temperature is $\left(T_{\text{initial }}\right)$ is $5.5° \mathrm{C}$.

The final temperature is$\left(T_{\text{final }}\right)$ is$64.8° \mathrm{C}$.

The heat lost can be converted from joules to calories using the conversion factor shown below.

Step 2:The tempature change (part a)

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

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

$={64.8}^{\circ } \mathrm{C}-{5.5}^{\circ } \mathrm{C}$

$=59.3° C$

The temperature change $\left(\Delta T\right)$ is $59.3° \mathrm{C}$.

Step 3: The heat equation (part a)

The heat equation is as follows.

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

By substituting the value in the preceding equation, we obtain

$\text{ Heat }=5.25 \mathrm{g}\times 59.3° \mathrm{C}\times 4.184 \mathrm{J}/\mathrm{g}°\mathrm{C}$

$=13.0\times {10}^2 \mathrm{J}$

As a result, the required heat is$13.0\times {10}^2 \mathrm{J}$

Step 4:The heat required can be converted from joules into calories (part a)

The heat lost can be converted from joules to calories using the conversion factor shown below.

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

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

The heat required is

$\text{ Heat }=13.0\times {10}^2 \mathrm{J}$

$=13.0\times {10}^2 \mathrm{J}\times \frac{1.00 \mathrm{cal}}{4.184 \mathrm{J} }$

$=311 Cal$

As a result, the required heat is $311 \mathrm{cal}$.

Step 5:Given data (part b)

(b) The mass of water is $75.0 \mathrm{g}$, and 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$86.4° \mathrm{C}$.

The final temperature is$\left(T_{\text{final }}\right)$ is $2.1° \mathrm{C}$.

The heat lost can be converted from joules to calories using the conversion factor shown below.

Step 6:The temperature change (part b)

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

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

$={2.1}^{\circ } \mathrm{C}-{86.4}^{\circ } \mathrm{C}$

$=-84.3° C$

Therefore, the temperature change is $-84.3° \mathrm{C}$.

Step 7:The heat equation (part b)

The heat equation is as follows.

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

By substituting the value in the preceding equation, we obtain

$\text{ Heat }=75.0 \mathrm{g}\times \left(-84.3° \mathrm{C}\right)\times 4.184 \mathrm{J}/\mathrm{g}°\mathrm{C}$

The minus sign denotes that heat is lost during the process.. Therefore, the heat lost is $265\times {10}^2 \mathrm{J}$.

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

The heat lost can be converted from joules to calories using the conversion factor shown below.:

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

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

The heat required is

Heat $=265\times {10}^2 \mathrm{J}$

$=265\times 10{ }^2\mathrm{I}\times \frac{1.00 \mathrm{cal}}{4.184 \mathrm{I}}$

$=633\times 10{ }^{\mathrm{\prime }}\mathrm{cal}$

As a result, the required heat is $633\times {10}^1 \mathrm{cal}$.

Step 9:Given data (part c)

(c) The mass of silver is $10.0 \mathrm{g}$, and the specific heat $\left(SH\right)$ for silver is$0.235 \mathrm{J}/\mathrm{g}{}^{\circ }\mathrm{C}$. The initial temperature is $\left(T_{\text{initial }}\right)$ is $112° \mathrm{C}$.

The final temperature is $\left(T_{\text{final }}\right)$ is $275° \mathrm{C}$.

The heat lost can be converted from joules to calories using the conversion factor shown below.

Step 10:The temperature change (part c)

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

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

$={275}^{\circ } \mathrm{C}-{112}^{\circ } \mathrm{C}$

$=163° C$

As a result, the required heat is $163° \mathrm{C}$.

Step 11:The heat equation (part c)

The heat equation is as follows.

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

By substituting the value in the preceding equation, we obtain

Heat $=10.0 \mathrm{g}\times 163° \mathrm{C}\times 0.235 \mathrm{J}/\mathrm{g}°\mathrm{C}$

$=385 J$

As a result, the required heat is $385 \mathrm{J}$

Step 12:The heat required can be converted from joules into calories (part c)

The heat lost can be converted from joules to calories using the conversion factor shown below.

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

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

The heat required is

$\text{ Heat }=383 \mathrm{J}$

$=383 J\times \frac{1.00 \mathrm{cal}}{4.184 \mathrm{J}}$

$=91.6 Cal$

As a result, the required heat is $91.6 \mathrm{cal}$.

Step 13:Given data (part d)

(d) The mass of gold is $18.0 \mathrm{g}$, and the specific heat$\left(\mathrm{SH}\right.$)for iron is $0.129 \mathrm{J}/\mathrm{g}{}^{\circ }\mathrm{C}$. The initial temperature is $\left(T_{\text{initial }}\right)$ is$224° \mathrm{C}$.

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

We must calculate the required heat 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:

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

$={118}^{\circ } \mathrm{C}-{224}^{\circ } \mathrm{C}$

$=-106° C$

Therefore, the temperature change is $-106° \mathrm{C}$

The heat equation is

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

By substituting the value in the preceding equation, we obtain

$\text{ Heat }=18.0 \mathrm{g}\times \left(-{106}^{\circ } \mathrm{C}\right)\times 0.129 \mathrm{J}/{\mathrm{g}}^{\circ }\mathrm{C}$

$=-246 J$

The minus sign indicates that the heat is lost in the process. Therefore, the heat lost is$246 \mathrm{J}$.

The heat lost can be converted from joules to calories using the conversion factor shown below

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

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

The heat lost is

Heat $=246 \mathrm{J}$

$=246 J\times \frac{1.00 \mathrm{cal}}{4.184 \mathrm{J}}$

= 588 cal

As a result, the required heat is 58.8 cal.