91Ó°ÊÓ

(a) The mass of water is$5.25\mathrm{g}$, and the specific heat $\left(SH\right)$for water is localid="1653052589831" $4.184\mathrm{J}/\mathrm{g}°\mathrm{C}$The initial temperature is $\left(T_{\text{initial}}\right)$is localid="1653052595971" $5.5°\mathrm{C}$.

The final temperature is$\left(T_{\text{final}}\right)$islocalid="1653052605136" $64.8°\mathrm{C}$.

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

The temperature change $\left(\mathrm{Δ}T\right)$can be calculated by using the following formula:

$\mathrm{Δ}T=T_{\text{final}}−T_{\text{initial}}$

localid="1653052613916" $={64.8}^{∘}\mathrm{C}−{5.5}^{∘}\mathrm{C}$

localid="1653052620116" $=59.3°C$

The temperature change $\left(\mathrm{Δ}T\right)$is localid="1653052626606" $59.3°\mathrm{C}$.

The heat equation is as follows.

Heat $=$mass $×\mathrm{Δ}T×SH$

By substituting the value in the preceding equation, we obtain

localid="1653052638163" $\text{Heat}=5.25\mathrm{g}×59.3°\mathrm{C}×4.184\mathrm{J}/\mathrm{g}°\mathrm{C}$

localid="1653052645235" $=13.0×{10}^2\mathrm{J}$

As a result, the required heat islocalid="1653052651623" $13.0×{10}^2\mathrm{J}$

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

$\text{Heat}=13.0×{10}^2\mathrm{J}$

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

$=311Cal$

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

(b) The mass of water is $75.0\mathrm{g}$, and the specific heat $\left(SH\right)$for water is localid="1653052717859" $4.184\mathrm{J}/\mathrm{g}{}^{∘}\mathrm{C}$The initial temperature is$\left(T_{\text{initial}}\right)$islocalid="1653052723774" $86.4°\mathrm{C}$.

The final temperature is$\left(T_{\text{final}}\right)$is localid="1653052729775" $2.1°\mathrm{C}$.

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

The temperature change $\left(\mathrm{Δ}T\right)$can be calculated by using the following formula:

$\mathrm{Δ}T=T_{\text{final}}−T_{\text{initial}}$

localid="1653052740468" $={2.1}^{∘}\mathrm{C}−{86.4}^{∘}\mathrm{C}$

localid="1653052747540" $=-84.3°C$

Therefore, the temperature change is localid="1653052754087" $-84.3°\mathrm{C}$.

The heat equation is as follows.

Heat$=$mass $×\mathrm{Δ}T×SH$

By substituting the value in the preceding equation, we obtain

localid="1653052765500" $\text{Heat}=75.0\mathrm{g}×\left(-84.3°\mathrm{C}\right)×4.184\mathrm{J}/\mathrm{g}°\mathrm{C}$

The minus sign denotes that heat is lost during the process.. Therefore, the heat lost is localid="1653052771541" $265×{10}^2\mathrm{J}$.

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

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

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

The heat required is

Heat $=265×{10}^2\mathrm{J}$

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

$=633×10{}^{\mathrm{′}}\mathrm{cal}$

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

(c) The mass of silver is $10.0\mathrm{g}$, and the specific heat $\left(SH\right)$for silver islocalid="1653052830136" $0.235\mathrm{J}/\mathrm{g}{}^{∘}\mathrm{C}$. The initial temperature is $\left(T_{\text{initial}}\right)$is localid="1653052824312" $112°\mathrm{C}$.

The final temperature is $\left(T_{\text{final}}\right)$is localid="1653052838424" $275°\mathrm{C}$.

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

The temperature change $\left(\mathrm{Δ}T\right)$can be calculated by using the following formula:

$\mathrm{Δ}T=T_{\text{final}}−T_{\text{initial}}$

localid="1653052848805" $={275}^{∘}\mathrm{C}−{112}^{∘}\mathrm{C}$

localid="1653052855023" $=163°C$

As a result, the required heat is localid="1653052861088" $163°\mathrm{C}$.

The heat equation is as follows.

Heat$=$mass$×\mathrm{Δ}T×SH$

By substituting the value in the preceding equation, we obtain

Heat localid="1653052871100" $=10.0\mathrm{g}×163°\mathrm{C}×0.235\mathrm{J}/\mathrm{g}°\mathrm{C}$

localid="1653052877473" $=385J$

As a result, the required heat is localid="1653052883588" $385\mathrm{J}$

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

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

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

The heat required is

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

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

$=91.6Cal$

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

(d) The mass of gold is $18.0\mathrm{g}$, and the specific heat$\left(\mathrm{SH}\right.$)for iron is localid="1653052968300" $0.129\mathrm{J}/\mathrm{g}{}^{∘}\mathrm{C}$. The initial temperature is $\left(T_{\text{initial}}\right)$islocalid="1653052974647" $224°\mathrm{C}$.

The final temperature is$\left(T_{\text{final}}\right)$is localid="1653052981118" $118°\mathrm{C}$.

We must calculate the required heat in joules and calories.

The temperature change $\left(\mathrm{Δ}T\right)$can be calculated by using the following formula:

$\mathrm{Δ}T=T_{\text{final}}−T_{\text{initial}}$

localid="1653052992535" $={118}^{∘}\mathrm{C}−{224}^{∘}\mathrm{C}$

localid="1653052999068" $=-106°C$

Therefore, the temperature change is localid="1653053006237" $-106°\mathrm{C}$

The heat equation is

Heat$=$mass$×\mathrm{Δ}T×SH$

By substituting the value in the preceding equation, we obtain

localid="1653053015039" $\text{Heat}=18.0\mathrm{g}×\left(−{106}^{∘}\mathrm{C}\right)×0.129\mathrm{J}/{\mathrm{g}}^{∘}\mathrm{C}$

localid="1653053025035" $=-246J$

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

localid="1653053054078" $1.00\mathrm{cal}=4.184\mathrm{J}$

localid="1653053062424" $\frac{1.00\mathrm{cal}}{4.184\mathrm{J}}$

The heat lost is

Heat localid="1653053073495" $=246\mathrm{J}$

localid="1653053087241" $=246J×\frac{1.00\mathrm{cal}}{4.184\mathrm{J}}$

= 588 cal

As a result, the required heat is 58.8 cal.