Question

An object with a mass of `160 g` is dropped into `880 mL` of water at `0^@C`. If the object cools by `120 ^@C` and the water warms by `5 ^@C`, what is the specific heat of the material that the object is made of?

Answer 1

`1004.35jkg^(-1)K^(-1)`

Explanation:

suppose,
the specific heat of the object is `S`
now,
we know,
`1ml` water= `1gm`
`:.880ml` water=`880gm`
`=0.88kg`
and we know,
heat loss= heat gain
`0.16Sxx(120-5)=0.88xx4200xx5`
`=>18.4S=18480`
`=>S=1004.35jkg^(-1)K^(-1)`