Chemistry, 20.11.2019 19:31 Masielovebug

Heat is added to a 200.-gram sample of h2o(s) to melt the sample at 0°c. then the resulting h2o ( ) is heated to a final temperature of 65°c. determine the total amount of heat required to completely melt the sample. heat is added to a 200.-gram sample of h2o(s) to melt the sample at 0°c. then the resulting h2o ( ) is heated to a final temperature of 65°c. determine the total amount of heat required to completely melt the sample.

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Answer from: thefandomarmy24

$\boxed{{\text{121}}{\text{.188 kJ}}}$ of energy is required to melt the sample completely.

Further explanation:

Heat of fusion is the amount of energy needed to melt a solid at its melting point. It is represented by $\Delta {H_{\text{f}}}$ .

Step 1: The ice at ${\text{0 }}^\circ {\text{C}}$ is converted to water at $0{\text{ }}^\circ {\text{C}}$ .

The formula to calculate the amount of energy is as follows:

${\text{Q}} = {\text{n}}\Delta {H_{\text{f}}}$ ...... (1)

Here,

Q is the amount of heat required.

n is the number of moles of water.

$\Delta {H_{\text{f}}}$ is the heat of fusion of ice.

The formula to calculate the moles of water is as follows:

${\text{Moles of water}}=\dfrac{{{\text{Given mass of water}}}}{{{\text{Molar mass of water}}}}$ ...... (2)

Substitute 200 g for the given mass of water and 18 g/mol for the molar mass of water in equation (2).

$\begin{aligned}{\text{Moles of water}}&=\frac{{200{\text{ g}}}}{{{\text{18 g/mol}}}}\\&= 11.11{\text{ mol}}\\\end{aligned}$

Substitute 11.11 mol for n and 6.01 kJ/mol for $\Delta\text{H}_\text{f}$ in equation (1).

$\begin{aligned}{{\text{Q}}_1}&= \left( {{\text{11}}{\text{.11 mol}}} \right)\left( {\frac{{6.01{\text{ kJ}}}}{{1{\text{ mol}}}}} \right)\\&= 66.77{\text{ kJ}}\\\end{aligned}$

Step 2: The temperature of water is raised from $0{\text{ }}^\circ {\text{C}}$ to $65{\text{ }}^\circ {\text{C}}$ .

The formula to calculate the heat energy of steam is as follows:

$\text{Q}=\text{mc}\Delta\text{T}$ …… (3)

Here,

Q is the amount of heat transferred.

m is the mass of substance.

c is the specific heat of substance.

$\Delta\text{T}$ is the temperature change.

The temperature change for the above change can be calculated as follows:

$\begin{aligned}\Delta\text{T}&=65\;^\circ\text{C}-0\;^\circ\text{C}\\&=65\;^\circ\text{C}\end{aligned}$

Substitute 200 g for m, $65\;^\circ\text{C}$ for $\Delta\text{T}$ and $4.186\;\text{j/g}^\circ\text{C}$ for c in equation (3).

$\begin{aligned}{{\text{Q}}_2} &= \left( {{\text{200 g}}} \right)\left( {{\text{4}}{\text{.186 J/g }}^\circ {\text{C}}} \right)\left( {{\text{65 }}^\circ {\text{C}}} \right)\\&= 54418{\text{ J}}\\\end{aligned}$

This energy is to be converted into kJ. The conversion factor for this is,

$1{\text{ J}} = {10^{ - 3}}{\text{ kJ}}$

Therefore the energy can be calculated as follows:

$\begin{aligned}{{\text{Q}}_{\text{2}}}&= \left( {54418{\text{ J}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ kJ}}}}{{1{\text{ J}}}}} \right)\\&= 54.418{\text{ kJ}}\\\end{aligned}$

The total amount of energy can be calculated as follows:

$\text{Q}=\text{Q}_1+\text{Q}_2$ ...... (4)

Here, Q is the total energy and $\text{Q}_1$ ,and $\text{Q}_2$ are the values of energies calculated in first and second step respectively.

Substitute 66.77 kJ for $\text{Q}_1$ and 54.418 kJ for $\text{Q}_2$ in equation (4).

$\begin{aligned}{\text{Q}} &= 66.77{\text{ kJ}} + 54.418{\text{ kJ}}\\&= {\text{121}}{\text{.188 kJ}}\\\end{aligned}$

Learn more:

What is the enthalpy of the given reaction? Find the enthalpy of decomposition of 1 mole of MgO:

Answer details:

Grade: Senior School

Chapter: Thermodynamics

Subject: Chemistry

Keywords: energy, Q1, Q2, Q, 54.418 kJ, 66.77 kJ, 121.88 kJ, m, c, temperature change, amount of heat, total energy.

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