Q. 3.53 - 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 needed to melt $50.0 g$ of ice at $0^{\circ }C$ and to warm the liquid to ${65}^{\circ }C$

b. Kilocalories released when $15.0 g$ of steam condenses at ${100}^{\circ }C$ and the liquid cools to $0^{\circ }C$

C. kilojoules needed to melt $24.0 g$ of ice at $0^{\circ }C$, warm the liquid to ${100}^{\circ }C$, and change it to steam at ${100}^{\circ }C$.

Step-by-Step Solution

Answer

(a) The Joules need to melt $50.0 g$ ice at $0^{\circ }C$ and to warm the liquid to ${65}^{\circ }C$ is $3.03\times {10}^4 J$

(b) The Kilocalories released when $15.0 g$ of steam at ${100}^{\circ }C$ and cools at $0^{\circ }C$ is $9.60 Kcal$

(c) The Kilojoules needed to melt $24.0 g$ of ice at $0^{\circ }C$ warms the liquid to ${100}^{\circ }C$ and change steam to ${100}^{\circ }C$ is $7.22 KJ$

Step1: Given Information (part a).

The following are the given information as,

Mass of ice $=50.0 g$

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

Temperature to warm the liquid$={65}^{\circ }C$

Step2: Find the Joules to warm the liquid (part a).

The heat of fusion of water is $334 J/g$, while the mass of ice is $50.0 g$. The following is the formula for calculating the energy required to melt $50.0 g$ of ice:

$$\begin{gathered} Heat change=mass\times heat of fusion \\ =50.0 g\times \frac{334 J}{g} \\ =16700 J \end{gathered}$$

The Heat Equation for the changes from $0^{\circ }C$ to ${65}^{\circ }C$ as:

$Heat=m\times ∆t\times SH$

$$\begin{gathered} =50.0 g\times ({65}^{\circ }C-0^{\circ }C)\times \frac{4.184 J}{g^{\circ }C} \\ =13598 J \end{gathered}$$

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

$16700 J+13598 J=30298 J$

The Amount of Energy will be $30298 J$

Step3: Given Information (part b).

The given information are as follows:

Mass of steam$=15.0 g$

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

Liquid cools to $0^{\circ }C$

Step4: The Kilocalories released when 15 . 0   g of steam at 100 ∘ C and cools at 0 ∘ C (part b).

The mass of ice is $15.0 g$, the heat of fusion of water is $540 cal/g$. So, the formula used to calculate the energy needed to $15.0 g$ melt of ice is as follows:

$$\begin{gathered} Heat change=mass\times heat of fusion \\ =15.0 g\times \frac{540 cal}{g} \\ =8.10\times {10}^{3 }cal \end{gathered}$$

Heat equation to cool the liquid from ${100}^{\circ }C$ to $0^{\circ }C$ as:

$Heat=m\times ∆t\times SH$

$$\begin{gathered} =15.0 g\times ({100}^{\circ }C-0^{\circ }C)\times \frac{1 cal}{g^{\circ }C} \\ =1.50\times {10}^3 J \end{gathered}$$

The total energy is calculated by adding the energy required to melt and the energy required to warm the liquid. The following is how it's calculated:

$$\begin{gathered} 8.10\times {10}^3\mathrm{cal}+1.50\times {10}^3\mathrm{cal}=9.60\times {10}^3\mathrm{cal}\times \frac{0.001\mathrm{kcal}}{1\mathrm{cal}} \\ =9.60 Kcal \end{gathered}$$

The Energy will be $9.60 Kcal$

Step5: Given Information (part c).

The given information are as follows:

Mass of ice$=24.0 g$

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

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

Step6: The Kilojoules needed to melt 24 . 0   g of ice at 0 ∘ C and warms it to 100 ∘ C and changes it to steam at 100 ∘ C (part c).

The mass of ice is $24.0 g$, the heat of fusion of water is $334 J/ g$. So, the formula used to calculate the energy needed to melt $24.0 g$ of ice is as follows:

$$\begin{gathered} Heat change=mass\times heat of fusion \\ =24.0 g\times \frac{334 J}{g} \\ =8.02\times {10}^{3 }cal \end{gathered}$$

The heat to warm liquid from $0^{\circ }C$ to ${100}^{\circ }C$ is,

$Heat=m\times ∆t\times SH$

$$\begin{gathered} =24.0 g\times ({100}^{\circ }C-0^{\circ }C)\times \frac{4.184 J}{g^{\circ }C} \\ =1.00\times {10}^4 J \end{gathered}$$

The vaporization heat is $2260 J/g$. The formula for calculating the amount of energy required to evaporate is,

$$\begin{gathered} Heat=mass\times heat of vaporization \\ =24.0 g\times \frac{2260 J}{g} \\ =5.42\times {10}^4 J \end{gathered}$$

The total energy is equal to the sum of the energy required to melt and the energy required to warm the liquid. It's computed like this:

$$\begin{gathered} 8.02\times {10}^3 \mathrm{J}+1.00\times {10}^4 \mathrm{J}+5.42\times {10}^4 \mathrm{J}=7.22\times {10}^4 \mathrm{J}\times \frac{1 \mathrm{kJ}}{1000 \mathrm{J}} \\ =7.2 \mathrm{KJ} \end{gathered}$$

The Heat energy will be $7.22 KJ$