Question

An object with a mass of `21 g` is dropped into `400 mL` of water at `0^@C`. If the object cools by `150 ^@C` and the water warms by `12 ^@C`, what is the specific heat of the material that the object is made of?

Answer 1

The specific heat is `=6.38 kJkg^-1K^-1`

Explanation:

The heat is transferred from the hot object to the cold water.

For the cold water, `Delta T_w=12ºC`

For the object `DeltaT_o=150ºC`

`m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)`

The specific heat of water `C_w=4.186kJkg^-1K^-1`

Let, `C_o` be the specific heat of the object

The mass of the object is `m_o=0.021kg`

The mass of the water is `m_w=0.4kg`

`0.021*C_o*150=0.4*4.186*12`

`C_o=(0.4*4.186*12)/(0.021*150)`

`=6.38 kJkg^-1K^-1`