An object with a mass of #32 g# is dropped into #250 mL# of water at #0^@C#. If the object cools by #16 ^@C# and the water warms by #3 ^@C#, what is the specific heat of the material that the object is made of?

Question

An object with a mass of #32 g# is dropped into #250 mL# of water at #0^@C#.  If the object cools by #16 ^@C# and the water warms by #3   ^@C#, what is the specific heat of the material that the object is made of?

Lily Adams · Accepted Answer

Given
#m_o->"Mass of the object"=32g# #v_w->"Volume of water object"=250mL# #Deltat_w->"Rise of temperature of water"=3^@C# #Deltat_o->"Fall of temperature of the object"=16^@C# #d_w->"Density of water"=1g/(mL)# #m_w->"Mass of water"#
#=v_wxxd_w=250mLxx1g/(mL)=250g# #s_w->"Sp.heat of water"=1calg^"-1"""^@C^-1# #"Let "s_o->"Sp.heat of the object"# Now using the principle of calorimetry Heat gained by water equals heat lost by an object. #=>m_o xx s_o xxDeltat_o=m_wxxs_wxxDeltat_w# #=>32xxs_o xx16=250xx1xx3# #=>s_o=(250xx3)/(32xx16)# #~~1.46calg^"-1"""^@C^-1#

Xavier Adame · Answer

undefined Use the formula: $ Q = mc\Delta T $

$ Q_{object} = -Q_{water} $

$ c_{object} = \frac{{m_{object} \cdot \Delta T_{object}}}{{m_{water} \cdot \Delta T_{water}}} $

$ c_{object} = \frac{{32 \, \text{g} \cdot (-16 \, ^\circ \text{C})}}{{250 \, \text{g} \cdot 3 \, ^\circ \text{C}}} $