Question

An object with a mass of `50 g` is dropped into `300 mL` of water at `0^@C`. If the object cools by `80 ^@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=2508Jcolor(white)(l)kg^(-1)color(white)(l)K^(-1)`

Explanation:

`DeltaE=mcDeltatheta`, where:

• `DeltaE` = change in energy (`J`)
• `m` = mass of the object (`kg`)
• `c` = specific heat capacity of the object (`J` `kg^(-1)` `K^(-1)`)
• `Deltatheta` = change in temperature (`K`)

Assuming all heat is transferred we have:
`m_oC_oDeltatheta_o=m_wc_wDeltatheta_w`

`m_o=50g=0.05kg`
`Deltatheta_o=80^circC=80K`
`rho_w~~1000kgcolor(white)(l)m^(-3)`
`V_w=300mL=0.3L=0.0003m^3`
`m_w=1000*0.0003=0.3kg`
`Deltatheta_w=8^circC=8K`
`c_w~~4180Jcolor(white)(l)kg^(-1)color(white)(l)K^(-1)`

`0.05C_o*80=0.3*4180*8`

`C_o=(0.3*4180*8)/(0.05*80)=2508Jcolor(white)(l)kg^(-1)color(white)(l)K^(-1)`