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Mumz [18]
3 years ago
15

what is the specific heat of a substance that requires 99,100J of thermal energy to heat 3.47 kg of this substance from 15 C to

41C​
Physics
1 answer:
Nady [450]3 years ago
7 0

Answer:

1098.42 J/kg.c°

Explanation:

Q=mcT

you have : Q = 99100

m=3.47kg

and the change in temperature T = 41-15=26c°

Solve for specific heat c :

c = Q/mT === 99100/3.47(26) = 1098.42J/kg . c°

I hope that it's a clear solution and explanation.

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