Question

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​

Answer 1

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.