A 2267 g sample of a substance is initially at 285 C After a

A 22.67 g sample of a substance is initially at 28.5 °C. After absorbing 2689 J of heat, the temperature of the substance is 168.4 °C. What is the specific heat (c) of the substance?

Solution

The amount of heat (Q) is given by the formula :

Q = m . c . delta T

where m is the mass of substance in grams

c is the specific heat

and delta T is the change in temperature

So putting all the values we get :

2689 = 22.67 x c x (168.4 - 28.5)

c = 0.848 J/g^oC