The mass of an object is Adding of heat will raise the temperature of the object by What is the specific heat of the object?5.60 J/kg ∙ K1400 J/kg ∙ K0.00180 J/kg ∙ K3,500,000 J/kg ∙ K

m = mass = 250g = 0.25 kg
ΔQ = 14 J
ΔT = 10°C
k= specific heat
Use the formula:
ΔQ= kmΔT
Isolate k
k= ΔQ / mΔT
Replacing with the values given:
k = 14 J / ( 0.25 kg* 10°c ) = 5.6 j/kgK