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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 15°C to 41°C?

Respuesta :

Answer=1098.43 J / Kg K, working shown in photo
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Answer:1100 J/kg·K

Explanation:

Rationale:

99,100 J = 3.47 kg · C · (41 - 15)C = 840 J/kg·K

but it marked the for-mentioned answer of 1100 J/kg·K choice correct

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