what is the specific heat of a 25.0 g substance that absorbs 493.4 j and raises the temperature from 12.0°c to 34°c?

Specific heat of substance  = 0.897 J/g.°C

Explanation:

Given data:

Specific heat of substance = ?

Mass of substance = 25.0 g

Heat absorbed = 493.4 J

Initial temperature = 12.0 °C

Final temperature = 34°C

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 34°C -12.0°C

ΔT = 22°C

493.4 J  = 25 g ×c× 22°C

493.4 J  = 550 g.°C ×c

c = 493.4 J  / 550 g.°C

c = 0.897 J/g.°C

sorry i was late, but i don't know what assignment your talking about, i'm in flvs as well, im in earth space science. but u are behind pace bc every class should be in the 3's or 4'