Question

If 2050 J of heat are added to a 150 g object its temperature increases by 15°C.

(a) What is the heat capacity of this object?
(b) What is the object's specific heat?

Answer 1

When an object gets heated by a temperature ΔT energy needed, E = mcΔT

Here energy is given E = 2050 J

Mass of object = 150 g

Change in temperature ΔT = 15
`^0C` = 15 K

a) Heat capacity of an object equal to the ratio of the heat added to (or removed from) an object to the resulting temperature change.

So heat capacity = E/ΔT = 2050/15 = 136.67 J/K

b) We have E = mcΔT

c =
`(2050)/(150*10^(-3)*15)  = 911.11 J/kgK`

So object's specific heat = 911.11 J/kgK