Question

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 1

Answer:
`\Large\boxed{0.84~J/g^\circ C}`

Step-by-step explanation:

Given information

`Q~(Energy)=99,100~J`

`m=3.47~kg`

`\Delta T=(T_(Final)-T_(Initial))=(45-11)=34^\circ C`

Given formula

`Q=m* c*\Delta T`

`Q=Energy\\m=mass\\c=specific~heat\\\Delta T=Change~in~temperature`

Convert the unit of mass into Grams

`1~kg=1000~g\\`

`3.47~kg=3.47* 1000=3470~g`

Substitute values into the given formula

`(99100)=(3470)* c *(34)`

Simplify by multiplication

`99100=117980~c`

Divide 117980 on both sides

`99100/117980=117980~c/117980`

`\Large\boxed{c=0.84~J/g^\circ C}`

Hope this helps!! :)

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