An object with a mass of #120 g# is dropped into #800 mL# of water at #0^@C#. If the object cools by #24 ^@C# and the water warms by #3 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
ali ergin
Feb 7, 2016

#c_m=0,0008bar3#

Explanation:

#"800 m L of water"=800 mg=0,8 g#
#Q_w=m_w *c_w * Delta t_w " Heat gained by water"#
#Q_w=0,8*1*3=2,4 Cal#
#Q_m=m_m*c_m*Delta t_m" Heat given by mass"#
#Q_m=120*c_m*24#
#Q_m=2880*c_m#
#Q_m=Q_w#
#2880*c_m=2,4#
#c_m=(2,4)/2880#
#c_m=0,0008bar3#

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