Verified answer

De acordo com a compreensão de calorimetria, temos que a quantidade de calor necessária para liquefazer a o gelo  é de 440 kcal.

Resolução:

• Elevando a temperatura do gelo de -10 a 0º C:

[tex]\Large\displaystyle\begin{cases} \sf Q = \: ? \: cal \\\sf m = 5 \ kg \times 1000= \bf 5000 \: g \\\sf c_{gelo} = 0{,} 5 \: cal/g\cdot ^\circ C \\\sf T= 0 ^\circ C \\\sf T_ 0 = -10 ^\circ C \end{cases}[/tex]

Calculando:

[tex]\Large\displaystyle\text{${ \sf Q = m \cdot c \cdot (T-T_0)}$}\\\\\Large\displaystyle\text{${ \sf Q = 5000 \cdot 0{,}5 \cdot [0-(-10)]}$}\\\\\Large\displaystyle\text{${ \sf Q = 2500 \cdot [0+10]}$}\\\\\Large\displaystyle\text{${ \sf Q = 2500 \cdot 10}$}\\\\\Large\displaystyle\boxed{ \sf Q = 25000 \: cal }[/tex]

• Fundindo o gelo:

[tex]\Large\displaystyle\begin{cases} \sf Q = \: ? \: cal \\\sf m = 5 \ kg \times 1000= \bf 5000 \: g \\\sf L_{fus\tilde{a}o} = 80 \: cal/g\end{cases}[/tex]

Calculando:

[tex]\Large\displaystyle\text{${ \sf Q = m \cdot L }$}\\\\\Large\displaystyle\text{${ \sf Q = 5000 \cdot 80 }$}\\\\\Large\displaystyle\boxed{ \sf Q = 400000 \: cal }[/tex]

• Elevando a temperatura da água de 0 a 3º C:

[tex]\Large\displaystyle\begin{cases} \sf Q = \: ? \: cal \\\sf m = 5 \ kg \times 1000= \bf 5000 \: g \\\sf c_{agua } = 1 \: cal/g\cdot ^\circ C \\\sf T= 3 ^\circ C \\\sf T_ 0 = 0^\circ C \end{cases}[/tex]

Calculando:

[tex]\Large\displaystyle\text{${ \sf Q = m \cdot c \cdot (T-T_0)}$}\\\\\Large\displaystyle\text{${ \sf Q = 5000 \cdot 1 \cdot (3-0)}$}\\\\\Large\displaystyle\text{${ \sf Q = 5000 \cdot 3}$}\\\\\Large\displaystyle\boxed{ \sf Q = 15000 \: cal }[/tex]

• A quantidade de calor total será:

[tex]\Large\displaystyle\text{${ \sf Q_{total} = 25000+400000+15000}$}\\\\\Large\displaystyle\text{${ \sf Q_{total} = 440000 \: cal \Rightarrow \div 1000}$}\\\\\Large\displaystyle\boxed{ \sf Q_{total} = 440 \: kcal}[/tex]

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