An object with a mass of #30 g# is dropped into #750 mL# of water at #0^@C#. If the object cools by #15 ^@C# and the water warms by #80 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
ali ergin
Mar 27, 2016

#c_o=0,0075#

Explanation:

#"The heat energy given by a mass of 30 g:"#
#Q_1=m_o*c_o*Delta t_o#
#m_o:" mass of object"#
#c_o:"specific heat of object"#
#Delta t_o:"change of temperature of object"#
#Q_1=30.c_o*15#
#Q_1=450*c_o#
#"The heat energy taken by water:"#
#Q_2=m_w*c_w*Delta t_w#
#m_w:"mass of water"#
#c_w:"specific heat of water"#
#Delta t_w:"change of temperature of water"#
#m_w=d*V=1*750=750 g#
#Q_2=750*1*80#
#Q_2=60000" Cal"#

#Q_1=Q_2#
#450*c_o=60000#
#c_o=450/60000#

#c_o=0,0075#

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