Question

An object with a mass of `100 g` is dropped into `500 mL` of water at `0^@C`. If the object cools by `20 ^@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 the object is `1`

Explanation:

In such cases, heat lost is equal to heat gained. Further heat lost / gained is given by `mst`, where `m` is the mass of the object, `s` is its specific heat and `t` is rise / fall in temperature.

Hence let `s` be the specific heat of the object, then heat lost by the object is `100xxsxx20=2000s` calories

annd heat gained by water is `500xx1xx4=2000` calories.

Hence `2000s=2000` or `s=1`