Question

An object with a mass of `60 g` is dropped into `900 mL` of water at `0^@C`. If the object cools by `40 ^@C` and the water warms by `16 ^@C`, what is the specific heat of the material that the object is made of?

Answer 1

Specific heat of the material that the object is `6`.

Explanation:

In such cases heat lost by one equals heat gained by other. Heat lost / gained is given by product of mass `m`, specific heat `s` (which is `1` for water) and change in temperature,

Let the specific heat of the mass of `60g`, which cools by `40^oC` be `s`. Hence heat lost is `60xx40xxs` calories.

900mL of water means `900g` and it warms by 16∘C. Hence heat gained by it is `900xx16=14400` calories.

Hence `60xx40xxs=14400` or `x=14400/(60xx40)=6`

Hence, specific heat of the material that the object is `6`.