Chapter 13, Problem 56AE

Interpretation Introduction

Interpretation:

Energy needed for conversion of $100\text{ g}$ of ice to water at $20°\text{C}$ has to be calculated.

Concept Introduction:

Heat required by $\text{1 g}$ of a substance to increase the temperature by $1°\text{C}$ is termed as specific heat capacity of that substance. Energy absorbed or released by the substance is calculated as follows:

  $\text{Energy}=\left(\text{mass}\right)\left(\text{specific heat}\right)\left(\text{change in temperature}\right)$

Heat of fusion is amount of heat required to convert solid to liquid. Every substance has different heat of fusion. Heat of fusion for ice is $384\text{ J/g}$. It is represented as ${\text{ΔH}}_{\text{fus}}$ and calculated as follows:

  $\text{q}=\left(\text{mass}\right)\left({\text{ΔH}}_{\text{fus}}\right)$

Answer to Problem 56AE

Energy needed for conversion of $100\text{ g}$ of ice to water at $20°\text{C}$ is $48.78\text{ kJ}$.

Explanation of Solution

The formula to calculate energy of ice is as follows:

  $\text{E}=\text{m}\cdot \text{c}\cdot \left({\text{T}}_{\text{2}}-{\text{T}}_{\text{1}}\right)$        (1)

Here,

$\text{E}$ is energy of ice

$\text{m}$ is mass of ice

$\text{c}$ is specific heat of ice

${\text{T}}_{\text{2}}-{\text{T}}_{\text{1}}$ is change in temperature

Substitute $100\text{ g}$ for $\text{m}$, $2.01\text{ J/g }°\text{C}$ for $\text{c}$, $0°\text{C}$ for ${\text{T}}_{\text{2}}$ and $-10.0°\text{C}$ for ${\text{T}}_1$ in equation (1).

  $\begin{array}{c}\text{E}=\left(100\text{ g}\right)\left(2.01\text{ J/g }°\text{C}\right)\left(0°\text{C}-\left(-10°\text{C}\right)\right)\\ =\left(100\text{ g}\right)\left(\frac{2.01\text{ J}}{1\text{ g }°\text{C}}\right)\left(\frac{{10}^{-3}\text{ kJ}}{1\text{ J}}\right)\left(10°\text{C}\right)\\ =2.010\text{ kJ}\end{array}$

The formula to calculate energy needed for conversion of ice to water is as follows:

  $\text{q}=\text{m}\cdot {\text{ΔH}}_{\text{fus}}$        (2)

Here,

$\text{q}$ is the energy needed for conversion of ice to water

$\text{m}$ is the mass of ice

${\text{ΔH}}_{\text{fus}}$ is the heat of fusion

Substitute $100\text{ g}$ for $\text{m}$ and $384\text{ J/g}$ for ${\text{ΔH}}_{\text{fus}}$ in equation (2).

  $\begin{array}{c}\text{q}=\left(100\text{ g}\right)\left(\frac{384\text{ J}}{1\text{ g}}\right)\left(\frac{{10}^{-3}\text{ kJ}}{1\text{ J}}\right)\\ =38.400\text{ kJ}\end{array}$

The formula to calculate energy of water is as follows:

  $\text{E'}=\text{m}\cdot \text{c}\cdot \left({\text{T}}_{\text{2}}-{\text{T}}_{\text{1}}\right)$        (1)

Here,

$\text{E'}$ is energy of water

$\text{m}$ is mass of water

$\text{c}$ is specific heat of water

${\text{T}}_{\text{2}}-{\text{T}}_{\text{1}}$ is change in temperature

Substitute $100\text{ g}$ for m, $4.184\text{ J/g }°\text{C}$ for c, $20°\text{C}$ for ${\text{T}}_{\text{2}}$ and $0°\text{C}$ for ${\text{T}}_1$ in equation (1).

  $\begin{array}{c}\text{E'}=\left(100\text{ g}\right)\left(4.184\text{ J/g }°\text{C}\right)\left(20°\text{C}-\left(-0°\text{C}\right)\right)\\ =\left(100\text{ g}\right)\left(\frac{4.184\text{ J}}{1\text{ g }°\text{C}}\right)\left(\frac{{10}^{-3}\text{ kJ}}{1\text{ J}}\right)\left(20°\text{C}\right)\\ =8.368\text{ kJ}\end{array}$

Therefore, total energy needed for conversion of ice to water is as follows:

  $\text{Total energy of ice}=\text{E}+\text{q}+\text{E'}$        (3)

Substitute $2.010\text{ kJ}$ for E, $38.400\text{ kJ}$ for q and $8.368\text{ kJ}$ for $\text{E'}$ in equation (3).

  $\begin{array}{c}\text{Total energy of ice}=2.010\text{ kJ}+38.400\text{ kJ}+8.368\text{ kJ}\\ =48.78\text{ kJ}\end{array}$

Hence, energy needed for conversion of $100\text{ g}$ of ice to water at $20°\text{C}$ is $48.78\text{ kJ}$.