Answer:
160 kJ of energy must be removed from the soft drink for it reaches [tex]25^{o}C[/tex]
Explanation:
Liquid:
[tex]Q_{1}= m.c.\Delta T[/tex]
From the given ,
mass "m" = 340gm
c= 4.18 [tex]J/g^{o}C[/tex]
[tex]\Delta T = 25^{o}C[/tex]
[tex]Q_{1}= 340g\times4.18\times25[/tex]
[tex]Q_{1}= 35,5300J\,\,or\,\,35.5kJ[/tex]
Freezing;
[tex]Q_{2}= m.H_{f}[/tex]
[tex]Q_{2}= m.H_{f}[/tex]
[tex]Q_{2}= 113,560\,\,or\,113.6kJ[/tex]
Solid:
[tex]Q_{3}= m.c.\Delta T[/tex]
[tex]Q_{3}= 340g\times2.1J/g^{0}C\times12[/tex]
[tex]Q_{3}= 8568J or 8.6kJ[/tex]
[tex]Q_{Total}=Q_{1}+Q_{2}+Q_{3}[/tex]
[tex]=35.5kJ+113.6kJ+8.6kJ=158.6kJ\,or\,160kJ[/tex]
Therefore, 160 kJ of energy must be removed from the soft drink for it reaches [tex]25^{o}C[/tex].
