Chapter 13, Problem 29PE

Interpretation Introduction

Interpretation:

Final temperature for the energy of $75\text{ g}$ of ice that is added to $1.5\text{ L}$ of water 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 29PE

Final temperature for the energy absorbed by $75\text{ g}$ of ice is $69.33°\text{C}$.

Explanation of Solution

The formula for calculation of energy absorbed by $75\text{ g}$ 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 absorbed by ice

$\text{m}$ is mass of ice

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

${\text{T}}_{\text{2}}$ is final temperature

${\text{T}}_{\text{1}}$ is initial temperature

Substitute $75\text{ g}$ for $\text{m}$, $2.03\text{ J/g }°\text{C}$ for $\text{c}$ and $0°\text{C}$ for ${\text{T}}_1$ in equation (1).

  $\begin{array}{c}\text{E}=\left(75\text{ g}\right)\left(2.03\text{ J/g }°\text{C}\right)\left({\text{T}}_{\text{2}}-\text{0}°\text{C}\right)\\ \text{=152}{\text{.25T}}_{\text{2}}\text{ J}\end{array}$

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 $75\text{ g}$ for $\text{m}$ and $333.55\text{ J/g}$ for ${\text{ΔH}}_{\text{fus}}$ in equation (2).

  $\begin{array}{c}\text{q}=\left(75\text{ g}\right)\left(\frac{333.55\text{ J}}{1\text{ g}}\right)\\ =25016.25\text{ J}\end{array}$

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

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

Substitute $\text{152}{\text{.25T}}_{\text{2}}\text{ J}$ for E and $25,016.25\text{ J}$ for q in equation (3).

  $\text{Total energy of ice}=\text{152}{\text{.25T}}_{\text{2}}\text{ J}+25,016.25\text{ J}$

Density of water is ${10}^3\text{ g/L}$ so mass of water is $1.5\times {10}^3\text{ g}$.

The formula to calculate energy lost by water is as follows:

  ${\text{E}}^{\prime }=-\left(\text{m}\cdot \text{c}\cdot \left({\text{T}}_{\text{2}}-{\text{T}}_{\text{1}}\right)\right)$        (4)

Here,

${\text{E}}^{\prime }$ is energy lost by water

$\text{m}$ is mass of water

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

${\text{T}}_{\text{2}}$ is final temperature

${\text{T}}_{\text{1}}$ is initial temperature

Substitute $1.5\times {10}^3\text{ g}$ for $\text{m}$, $2.03\text{ J/g }°\text{C}$ for $\text{c}$ and $75°\text{C}$ for ${\text{T}}_1$ in equation (4).

  $\begin{array}{c}{\text{E}}^{\prime }=-\left(\left(1.5\times {10}^3\text{ g}\right)\left(\frac{4.184\text{ J}}{1\text{ g }°\text{C}}\right)\left({\text{T}}_{\text{2}}-\text{75}\right)°\text{C}\right)\\ =-\left(\left(\text{6276 J}\right)\left({\text{T}}_{\text{2}}-\text{75}\right)\right)\end{array}$

Since total energy needed for conversion of ice to water is equal to energy lost by water. Hence it is expressed as follows:

  $\text{152}{\text{.25T}}_{\text{2}}\text{ J}+25,016.25\text{ J}=-\left(\left(\text{6276 J}\right)\left({\text{T}}_{\text{2}}-\text{75}\right)\right)$        (5)

Rearrange equation (5) for final temperature, ${\text{T}}_{\text{2}}$.

  $\begin{array}{c}\text{152}{\text{.25T}}_{\text{2}}\text{ J}+25,016.25\text{ J}=-\left({\text{6276T}}_{\text{2}}-228,375\right)\text{ J }\\ \left(\text{152}{\text{.25T}}_{\text{2}}+{\text{6276T}}_{\text{2}}\right)\text{ J}=\left(470700-25016.25\right)\text{ J}\\ \text{6428}{\text{.25T}}_2\text{ J}=\text{445683}\text{.75}\text{}\text{ J}\\ {\text{T}}_2=69.33°\text{C}\end{array}$

Hence, final temperature for the energy absorbed by $75\text{ g}$ of ice is $69.33°\text{C}$.