An object with a mass of #4 g# is dropped into #650 mL# of water at #0^@C#. If the object cools by #12 ^@C# and the water warms by #30 ^@C#, what is the specific heat of the material that the object is made of?

Use Heat and the First Law of Thermodynamics. We can use the relationship #Q = mcΔT# to relate the temperature changes of the object and the water to their masses, specific heats, and the amount of heat supplied to or absorbed by each.
#ΔT_o = (ΔQ_o)/(M_oC_o)# ===================> (1) Object

#ΔT_(H_2O)=(ΔQ_(H_2O))/(M_(H_2O)C_(H_2O))# ===========> (2) #H_2O#

Divide equation (1) into (2) knowing that #(ΔQ_o) = ΔQ_(H_2O)#

#(ΔT_o)/(ΔT_(H_2O))= (M_(H_2O)C_(H_2O))/(M_oC_o)#

Now what the mass of 650 ml of water? The mass of 650 ml of water is #=>.65 kg#. From a specific heat table
#C_(H_2O)=75.2 J/"mol" K#. Now we have all we need, just substitute and solve for #C_o#

#C_o = (ΔT_(H_2O))* (M_(H_2O)C_(H_2O))/(M_o*ΔT_o)#

#C_o = (30^oC*650 g*4.184 ((Jg)/c) ) / (4 g *12^oC)#

Now is a question calculating. That said I am a bit suspicious of the answer it is too large for a specific heat value...

As the quantities have been given in CGS units, therefore, answer has been worked out in same units.

Value used

Specific heat of water#=1cal//gm=4.19kJ//kg#

Let the specific heat of the material of object #=s_O#

Amount of heat exchanged is given as #DeltaQ=msDeltat#,
where #m# is the mass, #s# is the specific heat and #Deltat# is change in temperature.

Heat gained by #650mL# of water at #0^oC#,
assuming #1mL# of water #=1gm# of water.

#DeltaQ_(gai n ed)=msDeltat_w=650 times 1times 30=19500cal#

Simlarly heat lost by object is given as
#DeltaQ_(lost)=ms_ODeltat_O=4xxs_Oxx12=48s_O#

Since, #DeltaHeat_(lost)= Delta Heat_(gai n ed)#

#:. 48s_O=19500#
#implies s_O=19500/48#
or # s_O=406.25cal//gm#