Question

An object with a mass of `18 g` is dropped into `400 mL` of water at `0^@C`. If the object cools by `150 ^@C` and the water warms by `12 ^@C`, what is the specific heat of the material that the object is made of?

Answer 1

`=(7.4311J)/(g*C)`

Explanation:

1. Given the volume of water used as medium to cool down the object, its mass can be derived through the density formula; that is,
   `rho=m/V`
   `m=rhoxxV`
   `m=(1g)/cancel((ml))xx400cancel(ml)`
   `m=400gH_2O`
2. Find the heat absorbed (gained) by the water that increases its temperature by `12^o`. Knowing the `Cp_w=(4.18J)/(g*C)`, heat gained can be calculated by:
   `Q=mCpDeltaT`
   `Q=(400cancel(g))((4.18J)/cancel((g*C)))(12)cancel(^oC)`
   `Q=20,064J=20.064kJ`
3. Since heat lost=heat gained; therefore, the `Cp` of the material can be computed as;
   `Q=mCpDeltaT`
   `Cp_(o)=(Q)/(mDeltaT)`
   `Cp_(o)=(20,064J)/((18g)(150^oC))`
   `Cp_(o)=(7.4311J)/(g*C)`