Question

An object with a mass of `10 g` is dropped into `240 mL` of water at `0^@C`. If the object cools by `120 ^@C` and the water warms by `8 ^@C`, what is the specific heat of the material that the object is made of?

Answer 1

Here,the heat energy liberated by the object will be taken up by the water as a result of which,both will come to a thermal equilibrium.

So,heat energy liberated by the object is `10*s* 120` Calorie (using `H= msd theta`,where, `m` is the mass of the object, `s` is its specific heat and `d theta` is the change in temperature)

Now,heat energy absorbed by water is `240*1*1*8` calorie (where, mass of water = volume* density, and `1 ml =1 cm^3` and in CGS, density of water is `1 g/(cm^3)` and specific heat of water is `1 C g^-1 @C ^-1`)

So,equating both we get,

`1200s =1920`

so, `s= 1.6 C g^-1 @C^-1`

Answer 2

The specific heat is `=6.76.7 kJkg^-1K^-1`

Explanation:

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

For the cold water, `Delta T_w=8ºC`

For the object `DeltaT_o=120ºC`

`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=0.010kg`

The mass of the water is `m_w=0.24kg`

`0.010*C_o*120=0.24*4.186*8`

`C_o=(0.24*4.186*8)/(0.010*120)`

`=6.7 kJkg^-1K^-1`