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What is the specific heat of a substance that requires 99,100 J of thermal energy to heat 3.47 kg of this substance from 11°C to 45°C?

Answer :

Answer: [tex]\Large\boxed{0.84~J/g^\circ C}[/tex]

Explanation:

Given information

[tex]Q~(Energy)=99,100~J[/tex]

[tex]m=3.47~kg[/tex]

[tex]\Delta T=(T_{Final}-T_{Initial})=(45-11)=34^\circ C[/tex]

Given formula

[tex]Q=m\times c\times\Delta T[/tex]

[tex]Q=Energy\\m=mass\\c=specific~heat\\\Delta T=Change~in~temperature[/tex]

Convert the unit of mass into Grams

[tex]1~kg=1000~g\\[/tex]

[tex]3.47~kg=3.47\times 1000=3470~g[/tex]

Substitute values into the given formula

[tex](99100)=(3470)\times c \times(34)[/tex]

Simplify by multiplication

[tex]99100=117980~c[/tex]

Divide 117980 on both sides

[tex]99100\div117980=117980~c\div117980[/tex]

[tex]\Large\boxed{c=0.84~J/g^\circ C}[/tex]

Hope this helps!! :)

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Rewritten by : Barada