$\mathbf{180}\mathbf{\text{\hspace{0.17em}g}}$ of boiling water (temperature $\mathbf{100}{}^{\mathbf{0}}\mathbf{\text{C}}$, heat capacity $\mathbf{4}\mathbf{.2}\mathbf{\text{\hspace{0.17em}J}}\mathbf{/}\mathbf{\text{K}}\mathbf{/}\mathbf{\text{g}}$) are poured into an aluminum pan whose mass is 1050 g and initial temperature $\mathbf{26}{}^{\mathbf{0}}\mathbf{\text{C}}$(the heat capacity of aluminum is $\mathbf{0.9}\mathbf{\text{\hspace{0.17em}J}}\mathbf{/}\mathbf{\text{K}}\mathbf{/}\mathbf{\text{g}}$).

(a) After a short time, what is the temperature of the water?

(b) What simplifying assumptions did you have to make? (1) The thermal energy of the water doesn't change. (2) Thermal energy of the aluminum doesn't change. (3) Energy transfer between the system (water plus pan) and the surroundings was negligible during this time. (4) The heat capacities for both water and aluminum hardly change with temperature in this temperature range.

(c) Next you place the pan on a hot electric stove. While the stove is heating the pan, you use a beater to stir the water, doing 29541 J of work and the temperature of the water and pan increases to $\mathbf{86.9}{}^{\mathbf{0}}\mathbf{\text{C}}$. How much energy transfer due to a temperature difference was there from the stove into the system consisting of the water plus the pan?

The given data can be listed below as,

• The mass of boiling water is,${\mathrm{m}}_{\mathrm{b}}=180\text{\hspace{0.17em}g}$.
• The heat capacity of boiling water is${\mathrm{c}}_{\mathrm{w}}=4.2\text{\hspace{0.17em}J}/\text{K}/\text{g}$ .
• The temperature of boiling water is,${\mathrm{T}}_{\mathrm{iw}}=100{}^0\text{C}$.
• The mass of aluminium pan is,${\mathrm{m}}_{\mathrm{a}}=1050\text{\hspace{0.17em}g}$ .
• The initial temperature of aluminium pan is,${\mathrm{T}}_{\mathrm{ia}}=26{}^0\text{C}$.
• The heat capacity of aluminium pan is${\mathrm{c}}_{\mathrm{a}}=0.9\text{\hspace{0.17em}J}/\text{K}/\text{g}$ .
• The work done by water is, $\mathrm{W}=29541\text{\hspace{0.17em}J}$.
• The of water and pan increases by, $86.9{}^0\text{C}$.

Heat capacity is the ratio of heat absorbed to temperature change by a material. In terms of the actual amount of material being taken into account, which is most frequently a mole, it is typically stated as calories per degree (the molecular weight in grams). Specific heat is the heat capacity expressed in calories per gram.

According to the energy principle,

$\begin{array}{c}\text{Heatlostbywater}=\text{Heatgainedbyaluminium}\\ {\mathrm{m}}_{\mathrm{b}}{\mathrm{c}}_{\mathrm{w}}({\mathrm{T}}_{\mathrm{iw}}-{\mathrm{T}}_{\mathrm{f}})={\mathrm{m}}_{\mathrm{a}}{\mathrm{c}}_{\mathrm{a}}({\mathrm{T}}_{\mathrm{f}}-{\mathrm{T}}_{\mathrm{ia}})\end{array}$

Substitute the all value in the above equation.

$\begin{array}{c}180\text{\hspace{0.17em}g}\times 4.2\text{\hspace{0.17em}J}/\text{K}/\text{g}\times (373{}^{\text{0}}\text{C}-{\mathrm{T}}_{\mathrm{f}})=1050\text{\hspace{0.17em}g}\times 0.9\text{\hspace{0.17em}J}/\text{K}/\text{g}\times ({\mathrm{T}}_{\mathrm{f}}-299{}^{\text{0}}\text{C})\\ 756\times (373{}^{\text{0}}\text{C}-{\mathrm{T}}_{\mathrm{f}})=945\times ({\mathrm{T}}_{\mathrm{f}}-299{}^{\text{0}}\text{C})\\ {\mathrm{T}}_{\mathrm{f}}=332\text{\hspace{0.17em}K}\\ {\mathrm{T}}_{\mathrm{f}}=58.89{}^0\text{C}\end{array}$

Hence the temperature of water is, $58.89{}^0\text{C}$.

From part (a) we made the following two assumptions, the heat capacities for both water and aluminium hardly change with temperature in this temperature range and energy transfer between the system (water plus pan) and the surroundings was negligible during this time. This is due to the fact that the energy transmitted remains inside the system and the system's overall energy is conserved.

The total conserved thermal energy of the system is expressed as,

$\begin{array}{l}{\mathrm{E}}_{\mathrm{t}}={\mathrm{m}}_{\mathrm{a}}{\mathrm{c}}_{\mathrm{a}}\mathrm{\Delta T}+{\mathrm{m}}_{\mathrm{b}}{\mathrm{c}}_{\mathrm{w}}\mathrm{\Delta T}\\ {\mathrm{E}}_{\mathrm{t}}=({\mathrm{m}}_{\mathrm{a}}{\mathrm{c}}_{\mathrm{a}}+{\mathrm{m}}_{\mathrm{b}}{\mathrm{c}}_{\mathrm{w}})\mathrm{\Delta T}\end{array}$

Substitute all the value in the above equation.

$\begin{array}{l}{\mathrm{E}}_{\mathrm{t}}=(1050\text{\hspace{0.17em}g}\times 0.9\text{\hspace{0.17em}J}/\text{K}/\text{g}+180\text{\hspace{0.17em}g}\times 4.2\text{\hspace{0.17em}J}/\text{K}/\text{g})\times (86.9{}^{\text{0}}\text{C}-58.89{}^{\text{0}}\text{C})\\ {\mathrm{E}}_{\mathrm{t}}=47645.01\text{\hspace{0.17em}J}\end{array}$

The 1^st law of thermodynamics is expressed as,

$\mathrm{Q}={\mathrm{E}}_{\mathrm{t}}-\mathrm{W}$

Substitute all the value in the above equation.

$\begin{array}{l}\mathrm{Q}=47645.01\text{\hspace{0.17em}J}-29541\text{\hspace{0.17em}J}\\ \mathrm{Q}=18104.01\text{\hspace{0.17em}J}\end{array}$

Hence the energy transfer due to a temperature difference is,$18104\text{\hspace{0.17em}J}$ .