Question

An object with a mass of `4 g` is dropped into `650 mL` of water at `0^@C`. If the object cools by `12 ^@C` and the water warms by `30 ^@C`, what is the specific heat of the material that the object is made of?

Answer 1

`C_o = (ΔT_(H_2O))* (M_(H_2O)C_(H_2O))/(M_o*ΔT_o)`

`C_o = (30^oC*650 g*4.184 ((Jg)/c) ) / (4 g *12^oC)`

Explanation:

Use Heat and the First Law of Thermodynamics. We can use the relationship `Q = mcΔT` to relate the temperature changes of the object and the water to their masses, specific heats, and the amount of heat supplied to or absorbed by each.
`ΔT_o = (ΔQ_o)/(M_oC_o)` ===================> (1) Object

`ΔT_(H_2O)=(ΔQ_(H_2O))/(M_(H_2O)C_(H_2O))` ===========> (2) `H_2O`

Divide equation (1) into (2) knowing that `(ΔQ_o) = ΔQ_(H_2O)`

`(ΔT_o)/(ΔT_(H_2O))= (M_(H_2O)C_(H_2O))/(M_oC_o)`

Now what the mass of 650 ml of water? The mass of 650 ml of water is `=>.65 kg`. From a specific heat table
`C_(H_2O)=75.2 J/"mol" K`. Now we have all we need, just substitute and solve for `C_o`

`C_o = (ΔT_(H_2O))* (M_(H_2O)C_(H_2O))/(M_o*ΔT_o)`

`C_o = (30^oC*650 g*4.184 ((Jg)/c) ) / (4 g *12^oC)`

Now is a question calculating. That said I am a bit suspicious of the answer it is too large for a specific heat value...

Answer 2

Specific heat of material of object`=406.25cal//gm`

The value is unrealistically high.

Explanation:

As the quantities have been given in CGS units, therefore, answer has been worked out in same units.

Value used

Specific heat of water`=1cal//gm=4.19kJ//kg`

Let the specific heat of the material of object `=s_O`

Amount of heat exchanged is given as `DeltaQ=msDeltat`,
where `m` is the mass, `s` is the specific heat and `Deltat` is change in temperature.

Heat gained by `650mL` of water at `0^oC`,
assuming `1mL` of water `=1gm` of water.

`DeltaQ_(gai n ed)=msDeltat_w=650 times 1times 30=19500cal`

Simlarly heat lost by object is given as
`DeltaQ_(lost)=ms_ODeltat_O=4xxs_Oxx12=48s_O`

Since, `DeltaHeat_(lost)= Delta Heat_(gai n ed)`

`:. 48s_O=19500`
`implies s_O=19500/48`
or `s_O=406.25cal//gm`