Question

An object with a mass of `32 g` is dropped into `480 mL` of water at `0^@C`. If the object cools by `120 ^@C` and the water warms by `4 ^@C`, what is the specific heat of the material that the object is made of?

Answer 1

Specific heat of object is `0.50`.

Explanation:

Let the specific heat of object be `s`.

As quantum of heal gained / lost is given by `mxxsxx(t_2-t_1)`, where `m` is mass of object gaining / loosing heat and `s` is its specific heat and `(t_2-t_1)` is difference in temperature caused by heating / cooling,

heat lost by object is `32xxsxx120`

and heat gained by water is `480xx1xx4`.

Now heat lost should be equal to heat gained

`32xxsxx120=480xx1xx4`

i.e. `s=(480xx4)/(32xx120)=(cancel4xxcancel120xxcancel4)/(cancel4xxcancel4xx2xxcancel120)=1/3=0.50`

Hence specific heat of object is `0.50`.