# Calculate the specific heat of a substance of a 35g sample absorbs 48 J as the temperature is raised from 293 K to 313 K

1 answer

###### Question:

Calculate the specific heat of a substance of a 35g sample absorbs 48 J as the temperature is raised from 293 K to 313 K

## Answers

Answer: 0.0686 J/g ° CExplanation:The specific heat capacity is the amount of heat needed to raise 1 gram of a substance by 1° CWe are given temperatures in Kelvin; so we need to convert it to ° C ° C = K - 273.15For the initial temperature T₁ = 293 K T₁ = 293 - 273.15 ° CT₁ = 19.85 ° CFor the final temperature T₂T₂ = 313 - 273.15 ° CT₂ = 39.85 ° CChange in temperature = ΔT = T₂ - T₁ ΔT = 39.85 ° C - 19.85 ° C ΔT = 20.00 ° Cmass given = 35 gheat absorbed (q) = 48 Jspecific heat (c) = ????Using the expression:q = mcΔT we have:48 = 35 × c × 2048 = c × 700c = [tex]\frac{48}{700}[/tex]c = 0.0686 J/g ° C

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