Question

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?

Answer 1

`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`