An object with a mass of #21 g# is dropped into #250 mL# of water at #0^@C#. If the object cools by #160 ^@C# and the water warms by #12 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Jul 12, 2017

Answer:

#3733.0374Jkg^-1°C^-1#

Explanation:

#m=21g=0.021kg#
#V_w=250ml=250cm^3=250*(10^-2)^3m^3=2.5*10^-4m^3#
#T_(wi)=0°C#
#T_(wf)=12°C#
#Deltatheta_(m)=160C#

Symbols:
#m=#mass
#v=#volume
#T=#temperature
#Deltatheta=#change in temperature
#Q=#Heat
#c=#specific heat capacity

Subscript:
#w=#water
#i=#initial
#f=#final
#m=#object (a mass)

Heat lost from the object = Heat gained by the water
assuming no heat lost to surrounding
#DeltaQ_w=DeltaQ_m#

#Q=mcDeltatheta#
#m_wc_wDeltatheta=m_mc_mDeltatheta#

density of water#=1gcm^-3=1000kgm^-3#
mass of water,#m_w=1000*2.5*10^-4=0.25kg#
specific heat capacity of water, #c_w=4181Jkg^-1°C^-1#

#0.25*4181*12=0.021*c_m*160#
#12543=3.36*c_m#
#c_m=12543/3.36~~3733.0374Jkg^-1°C^-1#