Question

A 36.07 g sample of a substance is initially at 27.8°C. After absorbing 2639 J of heat, the temperature of the substance is 109.0°C. What is the specific heat of the substance?

Answer 1

I found: `c=0.9J/(g^@C)`

Explanation:

I would use the relationship between heat `Q`, mass `m`, specific heat `c` and change in temperature `DeltaT` as:
`Q=mcDeltaT`
so:
`2639=36.07*c*(109.0-27.8)`
rearranging:
`c=0.9J/(g^@C)`

Aluminum perhaps?