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?

Question

asked Dec 28, 2023 86.1k views

1 Answer

Sahithi · Answer 1 · 2024-01-01T18:14:53+0000

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!! :)

Please let me know if you have any questions