Q. 3.40 - An Introduction to General, Organic, and Biological Chemistry

Question

Use the heat equation to calculate the energy for each of the following (see Table$3.11$ ):

a. calories lost when$85 g$ a of waler cools from $45°\mathrm{C}$to $25°\mathrm{C}$

h. joules to heat$75 \mathrm{g}$ of water from $222°\mathrm{C}$to $66°\mathrm{C}$

C. kilocalories to heat $5.0 \mathrm{kg}$ of water from $22{}^{\circ }\mathrm{C}$ to $28{}^{\circ }\mathrm{C}$

d. kilojoules to heat $224 \mathrm{g}$of gold from$18°\mathrm{C}$ to $185°\mathrm{C}$

Step-by-Step Solution

Answer

(part a) As a result, the heat loss is$1700 cal$

(part b) As a result, the heat loss is$13807 J$

(part c) As a result, the heat loss is$30000 cal$

(part d) As a result, the heat loss is$4.83\times {10}^3 \mathrm{J}$

Step 1: Given data (part a)

a) The following is the heat equation for calculating energy:

Heat $=mass\left(m\right)\times$ temperature change$\left(\Delta T\right)\times$ specific heat $(S H)$

Mass of the water is $85 \mathrm{g}$.

The temperature fluctuates between$45° \mathrm{C}$ to $25° \mathrm{C}$

Step 2: Calculate heat lost (part a)

$\mathrm{\Delta }T={45}^{\circ } \mathrm{C}-{25}^{\circ } \mathrm{C}$

$=20° C$

Specific heat of water $=1 \mathrm{cal}/{\mathrm{g}}^{0 }\mathrm{C}$

Calculate the heat lost as follows:

Heat lost

$=85 \mathrm{g}\times 20° \mathrm{C}\times 1 \mathrm{cal}/\mathrm{g}°\mathrm{C}$

As a result, the heat is$1700Cal$

Step 3:Given data (part b)

(b)

The following is the heat equation for calculating energy:

Heat $=mass\left(m\right)\times$ temperature change $(∆T)\times$ specific heat $\left(SH\right)$

Mass of the water is $75 \mathrm{g}$.

 The temperature fluctuates between$22° C$ to $66° C$

Step 4: Calculate heat(part b)

$\mathrm{\Delta }T={66}^{\circ } \mathrm{C}-{22}^{\circ } \mathrm{C}$

$={44}^{\circ }\mathrm{C}$

Specific heat of water $=4.184 \mathrm{J}/{\mathrm{g}}^0\mathrm{C}$

Calculate the heat as follows:

$\text{ Heat }=75 \mathrm{g}\times 44° \mathrm{C}\times 4.184 \mathrm{J}/\mathrm{g}°\mathrm{C}$

$=13807 J$

As a result, the heat is$=13807 J$

Step 5:Given data(part c)

(C)

The following is the heat equation for calculating energy:

Heat $=mass\left(m\right)\times$temperature change $\left(\Delta T\right)\times$specific heat $\left(SH\right)$

Mass of the water is $5.0 \mathrm{kg}$

The temperature fluctuates between $22° C$ to $28° C$

Step 6:Calculate the heat(part c)

$\mathrm{\Delta }T={28}^{\circ }\mathrm{C}-{22}^{\circ } \mathrm{C}$

$=6° C$

Specific heat of water $=1 \mathrm{cal}/{\mathrm{g}}^0\mathrm{C}$

Calculate the heat as follows:

$\text{ Heat }=5000 \mathrm{g}\times 6° \mathrm{C}\times 1 \mathrm{cal}/\mathrm{g}{}^{\circ }\mathrm{C}$

$=30000 cal$

To calculate the heat, do the following:

$30000 \mathrm{cal}\times \frac{1 \mathrm{kcal}}{1000 \mathrm{cal}}=30 \mathrm{kca}l$

As a result, the heat is$3000 cal$

Step 7:Given data(part d)

d)The following is the heat equation for calculating energy:

Heat$=mass\left(m\right)\times$ temperature change$(∆T)\times$  specific heat$\left(SH\right)$

Mass of the water is $224 \mathrm{g}$.

The temperature fluctuates between $18° C$ to $185° C$

Step 8:Calculate the heat (part d)

$\mathrm{\Delta }T={185}^{\circ } \mathrm{C}-{18}^{\circ } \mathrm{C}$

$={167}^{\circ } \mathrm{C}$

Specific heat of gold $=0.129 \mathrm{J}/{\mathrm{g}}^0\mathrm{C}$

To calculate the heat, do the following:

$=4.83\times {10}^3 \mathrm{J}$

Convert the  calories to kilocalories

$4.83\times {10}^3 \mathrm{J}\times \frac{1 \mathrm{kJ}}{1000 \mathrm{J}}=4.83 \mathrm{kJ}$

As a result, the heat is$=4.83\times {10}^3 \mathrm{J}$