An object with a mass of #110 g# is dropped into #750 mL# of water at #0^@C#. If the object cools by #70 ^@C# and the water warms by #5 ^@C#, what is the specific heat of the material that the object is made of?
1 Answer
Answer:
Explanation:
We can find the specific heat of the material that the object is made of by using the first law of thermodynamics and calorimetry.
The first law of thermodynamics is given by the equation:
#color(crimson)(DeltaE_(th)=W+Q)# where
#E_(th)# is thermal energy,#W# is the work performed on/by a system and#Q# is heat transferred into/out of it
We can also express the thermal energy in terms of temperature change:
#color(skyblue)(DeltaE_(th)=McDeltaT)# where
#M# is the mass of the substance,#c# is its specific heat, and#DeltaT# is the change in temperature experienced by the substance
We can set
#color(indigo)(Q=McDeltaT)#
We are provided with the following information:

#>"M"=110"g"# 
#>"V"=750"mL"# 
#>(T_i)_("water")=0^o"C"# 
#>(T_f)_("water")=5^o"C"# 
#>(DeltaT)_("object")=70^o"C"#
It is known that the specific heat of water is
#750cancel("cm"^3)*(1"g")/(cancel("cm"^3))#
#=color(lightgreen)(750"g")#
We have enough information to calculate
#Q_("water")=M_("w")c_("w")DeltaT_("w")#
#=(750"g")(4.184"J"//"g"^o"C")(5^o"C")#
#=color(grey)(15690"J")#
Now, for the unknown object:
#Q_("object")=M_("o")c_("o")DeltaT_("o")#
Solving for
#c_("o")=Q_("o")/(M_"o"*DeltaT_"o")#
And since
#c_"o"=(15690"J")/(110"g"*70^o"C")#
#~~color(indigo)(2.038"J"//"g"^o"C")#