Ravi read that a biscuit packet gives 450 k cal of energy per 100 g. Now, he wanted to find the calorific value of some other substance. He took 0.1 kg of that substance as a sample and combusted it on a sample holder of a bomb calorimeter. Because of this, the temperature of 1 kg of ice present in the external chamber rose from `0^(@)C" to "50^(@)C`. What is the calorific value of the given sample ? It is given that the specific heat capacity of water `= 4200 J = 336000 J kg^(-1)`.

Explanation

The mass of paraffin oil is `m= 100 kg`.
The temperature change, `Delta t = (x-300)`
The amount of energy absorbed by paraffin oil `=4180xx10^(3)J`
The specific heat of paraffin oil is
Water equivalent M = MS (formula)
`M_(1)S_(1)=M_(2)S_(2)`
`S_(2)=(M_(1))/(M_(2))S_(1)`
`=(50)/(100)xx4180`
`=2090`
Specific heat capacity of paraffin oils `=2090 J kg^(-1)^(@)C^(-1)`.
`Deltat=(theta)/(mS)`
`=(4180xx10^(3))/(100xx2090)`
`=20`
`x=300xxDelta t =320 K`
Alternate method:
`4180xx10^(3)J` of energy will be 'x' kg of water. To raise its temperature by `20^(@)C,` i.e., water equivalent of paraffin oil.
`xrArrkg=(4180xx10^(3))/(4180xx20)=50 kg.`
Therefore, the water equivalent of the given paraffin oil is 50 kg.