An object with a mass of #50 g# is dropped into #300 mL# of water at #0^@C#. If the object cools by #80 ^@C# and the water warms by #8 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Dec 7, 2017

Answer:

#C_o=2508Jcolor(white)(l)kg^(-1)color(white)(l)K^(-1)#

Explanation:

#DeltaE=mcDeltatheta#, where:

  • #DeltaE# = change in energy (#J#)
  • #m# = mass of the object (#kg#)
  • #c# = specific heat capacity of the object (#J# #kg^(-1)# #K^(-1)#)
  • #Deltatheta# = change in temperature (#K#)

Assuming all heat is transferred we have:
#m_oC_oDeltatheta_o=m_wc_wDeltatheta_w#

#m_o=50g=0.05kg#
#Deltatheta_o=80^circC=80K#
#rho_w~~1000kgcolor(white)(l)m^(-3)#
#V_w=300mL=0.3L=0.0003m^3#
#m_w=1000*0.0003=0.3kg#
#Deltatheta_w=8^circC=8K#
#c_w~~4180Jcolor(white)(l)kg^(-1)color(white)(l)K^(-1)#

#0.05C_o*80=0.3*4180*8#

#C_o=(0.3*4180*8)/(0.05*80)=2508Jcolor(white)(l)kg^(-1)color(white)(l)K^(-1)#