Question

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

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 = $\frac{2050}{150*10^{-3}*15} = 911.11 J/kgK$

 So object's specific heat = 911.11 J/kgK