Question

An object with a mass of `21 g` is dropped into `250 mL` of water at `0^@C`. If the object cools by `160 ^@C` and the water warms by `12 ^@C`, what is the specific heat of the material that the object is made of?

Answer 1

`3733.0374Jkg^-1°C^-1`

Explanation:

`m=21g=0.021kg`
`V_w=250ml=250cm^3=250*(10^-2)^3m^3=2.5*10^-4m^3`
`T_(wi)=0°C`
`T_(wf)=12°C`
`Deltatheta_(m)=160C`

Symbols:
`m=`mass
`v=`volume
`T=`temperature
`Deltatheta=`change in temperature
`Q=`Heat
`c=`specific heat capacity

Subscript:
`w=`water
`i=`initial
`f=`final
`m=`object (a mass)

Heat lost from the object = Heat gained by the water
assuming no heat lost to surrounding
`DeltaQ_w=DeltaQ_m`

`Q=mcDeltatheta`
`m_wc_wDeltatheta=m_mc_mDeltatheta`

density of water`=1gcm^-3=1000kgm^-3`
mass of water,`m_w=1000*2.5*10^-4=0.25kg`
specific heat capacity of water, `c_w=4181Jkg^-1°C^-1`

`0.25*4181*12=0.021*c_m*160`
`12543=3.36*c_m`
`c_m=12543/3.36~~3733.0374Jkg^-1°C^-1`