91影视

The given data can be listed below as,

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

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,

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

Substitute the all value in the above equation.

$\begin{array}{c}180\text{鈥塯}脳4.2\text{鈥塉}/\text{K}/\text{g}脳(373{\text{鈥�}}^{\text{0}}\text{C}鈭�{\mathrm{T}}_{\mathrm{f}})=1050\text{鈥塯}脳0.9\text{鈥塉}/\text{K}/\text{g}脳({\mathrm{T}}_{\mathrm{f}}鈭�299{\text{鈥�}}^{\text{0}}\text{C})\\ 756脳(373{\text{鈥�}}^{\text{0}}\text{C}鈭�{\mathrm{T}}_{\mathrm{f}})=945脳({\mathrm{T}}_{\mathrm{f}}鈭�299{\text{鈥�}}^{\text{0}}\text{C})\\ {\mathrm{T}}_{\mathrm{f}}=332\text{鈥块}\\ {\mathrm{T}}_{\mathrm{f}}=58.89{\text{鈥�}}^0\text{C}\end{array}$

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

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,

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

Substitute all the value in the above equation.

$\begin{array}{l}{\mathrm{E}}_{\mathrm{t}}=(1050\text{鈥塯}脳0.9\text{鈥塉}/\text{K}/\text{g}+180\text{鈥塯}脳4.2\text{鈥塉}/\text{K}/\text{g})脳(86.9{\text{鈥�}}^{\text{0}}\text{C}鈭�58.89{\text{鈥�}}^{\text{0}}\text{C})\\ {\mathrm{E}}_{\mathrm{t}}=47645.01\text{鈥塉}\end{array}$

The 1^st law of thermodynamics is expressed as,

$\mathrm{Q}={\mathrm{E}}_{\mathrm{t}}鈭�\mathrm{W}$

Substitute all the value in the above equation.

$\begin{array}{l}\mathrm{Q}=47645.01\text{鈥塉}鈭�29541\text{鈥塉}\\ \mathrm{Q}=18104.01\text{鈥塉}\end{array}$

Hence the energy transfer due to a temperature difference is,$18104\text{鈥塉}$ .