An unknown substance with a mass of 100 grams absorbs 1000 J while undergoing a temperature increase of 15 degrees Celsius what is the specific heat of the substance

Answer 1

Equate 1000 Joules to (0.1 Kilograms) Times (c Joules Per Kilogram-Dot-Degree-Centigrade)

Times (15 Degrees-Centigrade).

That is, 1000 J = (0.1 kg)(c J/(kg • °C))(15°C).

Then Specific Heat or c is given as (1000/1.5) or 666.6666666 J/(kg • °C).

Answer 2

q = m c ΔT

q = heat absorbed = 1000 J

m = mass = 100 g

c = specific heat = ?

ΔT = 15 °C

c = q / (m ΔT)

= 1000 J / (100 g x 15 °C)

= 0.667 J g^-1 °C^-1 ~ 0.7 J g^-1 °C^-1 (rounded to 1 significant figure)