Question

An object with a mass of `70 g` is dropped into `700 mL` of water at `0^@C`. If the object cools by `36 ^@C` and the water warms by `8 ^@C`, what is the specific heat of the material that the object is made of?

Answer 1

`c_o=0,002`

Explanation:

`m=700 m g=0,7 g " mass of water"`
`Q_w=m*c*Delta t`
`Q_w=0,7*1*8=5,6 Cal " Heat gained by water"`
`Q_o=m_o*c_o*Delta t_o`
`Q_o=70* c_o*36`
`Q_w=Q_o`
`70*c_o*36=5,6`
`c_o=(5,6)/(70*36)`
`c_o=(5,6)/2520`
`c_o=0,002`