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prohojiy [21]
2 years ago
9

A 55.0 g aluminum block initially at 27.5 ∘c absorbs 725 j of heat. what is the final temperature of the aluminum?

Chemistry
1 answer:
Airida [17]2 years ago
8 0

Final temperature of the aluminum is 41.8 °C

We have the equation for energy

E = mc∆T

Here m = 55 g = 0.055 kg

∆T=T-27.5

Specific heat capacity of aluminum = 921.096 J/kg.K

E = 725 J

Substituting

E = mc∆T

725 0.055 x 921.096 x (T - 27.5)

T-27.5 14.31

T=41.81° C = 41.8 °C

41.8 °C

Final temperature of the aluminum = 41.8°C

Learn more about Temperature here:

brainly.com/question/25677592

#SPJ4

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\large \boxed{\text{-41.2 kJ/mol}}

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We can calculate the enthalpy change of a reaction by using the enthalpies of formation of reactants and products

\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)

(a) Enthalpies of formation of reactants and products

\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{CO(g)} & -110.5 \\\text{H$_{2}$O} & -241.8\\\text{CO$_{2}$(g)} & -393.5 \\\text{H$_{2}$(g)} & 0 \\\end{array}

(b) Total enthalpies of reactants and products

\begin{array}{ccr}\textbf{Substance} & \textbf{Contribution)/(kJ/mol})&\textbf{Sum} \\\text{CO(g)} & -110.5& -110.5 \\\text{H$_{2}$O(g)} &-241.8& -241.8\\\textbf{Total}&\textbf{for reactants} &\mathbf{ -352.3}\\&&\\\text{CO}_{2}(g) & -393.5&-393.5 \\\text{H}_{2} & 0 & 0\\\textbf{Total}&\textbf{for products} & \mathbf{-393.5}\end{array}

(c) Enthalpy of reaction \Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)= \text{-393.5 kJ/mol - (-352.3 kJ/mol}\\= \text{-393.5 kJ/mol + 352.3 kJ/mol} = \textbf{-41.2 kJ/mol}\\ \text{The total enthalpy change is $\large \boxed{\textbf{-41.2 kJ/mol}}$}

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