An object with a mass of #120 g# is dropped into #800 mL# of water at #0^@C#. If the object cools by #30 ^@C# and the water warms by #5 ^@C#, what is the specific heat of the material that the object is made of?

Given,volume of water #800 ml#,we know density of water is #1 (gm)/(cm^3)# ,so it's mass will be #800 gm#

Suppose,the specific heat of the object be #s#,then using law of thermal equilibrium we can write,

# 120×30×s = 800×5×1 #

Solving we get, #s = 1.11# CGS units

The important concept here is that #"Heat lost=Heat gained"#

Since the mass of the water is not given, we can derive it through its volume at a given temperature as provided. Knowing these, mass of the water can be calculated using the density formula; i.e.,

#Density(rho)=("mass"(m))/("Volume"(V))#
#m=rhoxxV#

where:
#V=800ml " at " 0^oC# https://hypertextbook.com/facts/2007/AllenMa.shtml

#m=(0.9998g)/cancel(ml)xx800cancel(ml)#
#m=799.84g#

When the object was submerged in the water, temperature of the water has increased to 5^o that signifies heat #(Q)# has been absorbed and that can be computed as follows:

#Q=mCpDeltaT#

where:

#m_(H_20)=799.84g#
#T_i=0^oC#
#T_f=5^oC#
#Cp_w=(4.18J)/(g*^oC)#

#Q=799.84cancel(g)xx(4.18J)/cancel((g*oC))xx5cancel(*^oC)#
#Q=16,716.66J#

Remember that #"Heat lost" = "Heat gained"=16,716.66J#. So that, heat capacity of the object #(Cp_o)# of this process can be computed with reference to this value of Q. Rearrange formula to isolate the #Cp_o#; that is,

#Q=mCpDeltaT#
#Cp=(Q)/(mDeltaT)#

where:

#m=120g#
#DeltaT=30^oC#

#Cp=(16,716.66J)/(120gxx30^oC#
#Cp=(4.64J)/(g*^oC)#

The heat is transferred from the hot object to the cold water.

For the cold water, # Delta T_w=5ºC#

For the object #DeltaT_o=30ºC#

# m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)#

The specific heat of water #C_w=4.186kJkg^-1K^-1#

Let, #C_o# be the specific heat of the object

The mass of the object is #m_o=0.12kg#

The mass of the water is #m_w=0.800kg#

#0.12*C_o*30=0.800*4.186*5#

#C_o=(0.800*4.186*5)/(0.12*30)#

#=4.65 kJkg^-1K^-1#