Question

An object with a mass of `120 g` is dropped into `720 mL` of water at `0^@C`. If the object cools by `24 ^@C` and the water warms by `3 ^@C`, what is the specific heat of the material that the object is made of?

Answer 1

Specific heat of object is `0.75`.

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 objece is `120xxsxx24`

and heat gained by water is `720xx1xx3`.

Now heat lost should be equal to heat gained

`120xxsxx24=720xx1xx3`

i.e. `s=(720xx3)/(120xx24)=(6cancel720xxcancel3)/(cancel120xx8cancel24)=6/8=3/4=0.75`

Hence specific heat of object is `0.75`.