Question

An object with a mass of `20 g` is dropped into `750 mL` of water at `0^@C`. If the object cools by `90 ^@C` and the water warms by `18 ^@C`, what is the specific heat of the material that the object is made of?

Answer 1

suppose,the specific heat of the object is `s Cg^-1^@C^-1`

So,during this process of achieving thermal equilibrium,the heat energy released by the object was taken by the water.

So,we can write,

`20*s*90 = 750*1*18` (using, `H=msd theta`,where, `d theta` is the change in temperature,and specific heat for water is `1` CGS unit and weight of `750mL` of water is `750g`)

So,we get, `s=7.5 Cg^-1^@C^-1`

Answer 2

The specific heat is `=31.4 kJkg^-1K^-1`

Explanation:

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

For the cold water, `Delta T_w=18ºC`

For the hot object `DeltaT_o=90ºC`

`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=0.020kg`

The mass of the water is `m_w=0.75kg`

`0.020*C_o*90=0.75*4.186*18`

`C_0=(0.75*4.186*18)/(0.020*90)`

`=31.4 kJkg^-1K^-1`