Question

An object with a mass of `160 g` is dropped into `880 mL` of water at `0^@C`. If the object cools by `18 ^@C` and the water warms by `8 ^@C`, what is the specific heat of the material that the object is made of?

Answer 1

`s_(object)=2.dot 4` `cal//gm`

Explanation:

Assumption : Specific gravity of water at `0^@ C` is 1. (actually 0.9998)
`=>` weight of `880 ml` of water is `880 gm`.

Now by Law of conservation of energy
Heat lost by object = Heat gained by water.

Also Heat gained or lost `=m*s*Delta t`
where `m` is the mass of object, `s` its specific heat and`Delta t` Change in the temperature of the object.

Heat lost by object`=160*s_(object)*18`
Heat gained by water `=880*1*8`
(Specific heat of water in cgs units is 1 calorie/gm)
Equating two we obtain
`160*s_(object)*18=880*1*8`
`s_(object)=(880*8)/(160*18)`
`s_(object)=2.dot 4` `cal//gm`